Computer techniques in power system analysis [3 ed.] 9789332901131, 9332901139

Table of contents :
Title
Chapter 1
1 Introduction
2 Power System Components
3 Topological Analysis of Power Networks
4 Bus Impedance Algorithm
5 Short Circuit Studies
6 Power Flow Studies
7 Optimum Power Flow
8 Stability Studies
9 Renewable sources-Wind Energy System
Appendix A
Appendix B
Appendix C
Appendix D
Reference
Index

Citation preview

COMPUTER TECHNIQUES in POWER SYSTEM ANALYSIS Third Edition

COMPUTER TECHNIQUES in POWER SYSTEM ANALYSIS Third Edition

M A Pai

USA

Dheeman Chatterjee

McGraw Hill Education (India) Private Limited NEW DELHI New Delhi

Published by McGraw Hill Education (India) Private Limited, P-24, Green Park Extension, New Delhi 110 016. Computer Techniques in Power System Analysis, 3e Copyright © 2014, by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. ISBN (13): 978-93-329-0113-1 ISBN (10): 93-329-0113-9 Managing Director: Kaushik Bellani Head—Higher Education (Publishing and Marketing): Vibha Mahajan Senior Publishing Manager (SEM & Tech. Ed.): Shalini Jha Editorial Executive: Koyel Ghosh Manager—Production Systems: Satinder S Baveja Assistant Manager—Editorial Services: Sohini Mukherjee Senior Manager—Production: P L Pandita Assistant General Manager—Higher Education (Marketing): Vijay Sarathi Assistant Product Manager: Tina Jajoriya Senior Graphic Designer—Cover: Meenu Raghav General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Script Makers, 19, A1-B, DDA Market, Paschim Vihar, New Delhi 110 063 and printed at **** Cover Designer:

CONTENTS Preface

vii

1.1 1.2 1.3 1.4 1.5

Changing Scenario of Power and Energy Systems 1.1 Physical System, Computation, and Controls 1.2 Representation of Power System Components 1.5 Nature and Scope of Power System Studies 1.7 Conclusion 1.9

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9

Introduction 2.1 Per Unit System 2.1 Synchronous Generator Representation 2.3 Transmission Lines 2.6 Transformers 2.11 Load Representation 2.17 Representation of FACTS Controllers 2.19 Representation of HVDC Transmission System 2.29 Conclusion 2.32 Problems 2.35

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Introduction 3.1 Oriented Graphs 3.1 Primitive Impedance and Admittance Matrices 3.3 System Graph for Transmission Network 3.5 Relevant Concepts in Graph Theory 3.5 Transmission Network Representations 3.20 Network Matrices 3.24 Network Reduction 3.37 Distribution Factors 3.39 Conclusion 3.39 Problems 3.39

Introduction 4.1 Partial Network 4.1 Addition of Link(s) 4.3 Addition of Branch(es) 4.5 Removal of Link(s) 4.8 Implementation of the Algorithm Special Cases 4.10 Conclusion 4.10 Problems 4.16

4.9

vi

5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

Contents

Introduction 5.1 Three-Phase Balanced Networks and Faults 5.2 Symmetrical Components 5.3 Sequence Impedances for Symmetrical Faults 5.6 Symmetrical Fault Calculations 5.8 Three-Phase Networks 5.14 Fault Analysis in the Phase Impedance Form 5.16 General Fault Representation in Phase Quantities 5.18 Symmetrical Component Analysis 5.23 Currents and Voltages under Faulted Condition in Terms of Symmetrical Components 5.27 Conclusion 5.36 Problems 5.36

Introduction 6.1 Mathematical Model 6.2 Gauss–Seidel Method 6.5 Newton’s Method 6.9 6.13 Load Flow Solution in Large-Scale Systems 6.27 Inclusion of FACTS Devices 6.31 Inclusion of HVDC Transmission 6.40 Conclusion 6.42 Problems 6.42

7.3 7.4 7.5

Introduction 7.1 Classical Method of Economic Load Dispatch Without Considering Losses 7.1 Economic Load Dispatch Considering Losses 7.5 Optimal Power Flow 7.7 Conclusion 7.11

8.1

Introduction 8.1

7.1 7.2

8.3 8.4 8.5 8.6 8.7

8.1 Equal Area Criteria for the Determination of Stability Condition of SMIB System 8.3 Mathematical Representation for use in Transient Stability Simulation 8.5 Simulation Methods 8.12 Inclusion of FACTS Controllers 8.30 Conclusion 8.33 Problems 8.33

Contents

9.1 9.2 9.3 9.4 9.5

vii

Introduction 9.1 Wind Power Generation System Model and Power System Representation 9.2 Inclusion of Wind Power in Load Flow 9.5 Inclusion of Wind Power in Stability Studies 9.11 Conclusion 9.16 Problems 9.17

A.1 A.2 A.3 A.4 A.5

Introduction A.1 Iterative Methods for Linear System of Equations A.2 Direct Method of Solving Linear System of Equations A.3 System of Nonlinear Equations A.5 Matrix Factorization A.8

B.1 B.2 B.3 B.4 B.5

Introduction B.1 Euler’s Method B.1 Runge–Kutta Method B.3 Higher-Order Differential Equations B.6 Conclusion B.7

C.1 C.2 C.3 C.4 C.5

IEEE 14-Bus Test System C.1 IEEE 30-Bus Test Data C.3 Data for IEEE 57-Bus System C.7 DFIG and Wind Turbine Data C.13 Free Software For Power System Analysis and Links for Download C.13

D.1 Chapter 2: Power System Components D.1 D.2 Chapter 3: Topological Analysis of Power Networks D.3 Chapter 4: Bus Impedance Algorithm D.3 D.4 Chapter 5: Short-Circuit Studies D.3 D.5 Chapter 6: Power Flow Studies D.4 D.6 Chapter 8: Stability Studies D.6 References Index

D.2

R.1–R.2 I.1–I.3

undergraduate class at IIT Kanpur. The content of this book has also been used for several short courses for industry, to convey the essential system concepts involved in power network analysis to EE and non-EE audiences. Portions of the book have been class tested at the University of Illinois in a senior-level elective course on “Operation and Control of Power Systems.” Increase in the demand for electricity is a constant feature across the globe. This necessitates newer sources of energy and larger and more complex transmission and distribution networks. In the last few decades, there have been numerous grid failures affecting the lives of millions of people. Hence, a detailed study is required for running the existing power systems and planning their extension. Academicians as well as practicing engineers are switching over to a study of the numerous problems associated with planning and operation of large interconnected power systems. A power system analysis and operation course deals with the functioning principles of the power system—component-wise and as a whole. However, computer-based simulation is almost mandatorily associated with all these analyses. An effort has been made in this book to bridge the gap between the physics of the power system and the simulation. If a systematic solution of the problems of power systems is clarity in modern power system analysis. This book is an effort in this direction. It seeks to build on the courses the student has already had in the area of circuit and system theory. An earlier exposition to a course in power system analysis is preferred, though not essential.

at the postgraduate level. The material can be comfortably covered in one semester. The set of exercises at the end of the book is meant to test the students’ understanding of the subject as well as to give them computational experience. The revision was undertaken with Dr. Dheeman Chatterjee as co-author in response to the demands of the changing power sector scenario. Development of power electronics technology suitable for use in the voltage and power levels of transmission and distribution has provided power system operators and planners a new tool for better management of the system. Application of power electronics adds direct current (HVDC) transmission comes out to be cheaper in spite of the large cost of terminal equipment when used for carrying power over a long distance. Flexible AC transmission system (FACTS) helps utilize the transmission infrastructure better by enabling operation closer to the limits. The depleting sources of fossil fuel and their environmental impacts are forcing decision-makers to look for alternate renewable energy sources. The share of wind power, the cheapest renewable source, is increasing exponentially in different parts

ix

Preface

of the globe including the US and India. The variable nature of the source poses questions on possible impacts on the power system. Hence, future power system analysis and planning must include FACTS, HVDC and renewable sources like wind. The present revision has taken up these as the need of the hour. The modeling of HVDC and FACTS has been included in Chapter 2 along with the modeling of other power system components. Subsequently, the techniques for inclusion of FACTS and 6 and 8 respectively. A new chapter (Chapter 9) on wind energy system has been included. This includes a brief discussion of the methods for inclusion of the wind turbine and induction generator (both singly excited and doubly fed) in power system

the economic load dispatch. A brief introduction to that has been provided as a new chapter (Chapter 7) to make the content more complete. A summary of the new features is given below: ∑ ∑ ∑ ∑

New chapter on Economic Load Dispatch New chapter on Renewable Sources—Wind Energy Systems Comprehensive updates on HVDC, FACTS and WP Refreshed solved examples and illustrations

∑ Detailed study of the nature of problems requiring computational techniques in power system planning and operation ∑ Clear and rigorous description of bus impedance algorithm, short-circuit ∑ machine systems with emphasis on numerical simulation ∑ Test data of IEEE standard systems for ready reference ∑ Rich pedagogy includes ∑ Illustrations: 134 ∑ Unsolved numerical problems: 49 ∑ Solved examples: 33 There are nine chapters in the book. A brief chapter-wise overview is as follows. Chapter 1 gives an overview of the nature of problems arising during the planning and operation of power systems, which require computational techniques to solve. Chapter 2 provides the terminal relations of the components of modern power systems including generators, transformers, transmission lines (HVAC and HVDC) and FACTS controllers. Chapter 3 is devoted to an elementary exposition to linear graph theory and its use in deriving the multi-terminal representation of the transmission network. Chapter 4 contains the discussion of the algorithm for developing the bus impedance matrix of the transmission network based on graph theory. Chapter 5 describes short circuit analysis using the network representations where both phase impedance form and symmetrical component formulations are presented.

x

Computer Techniques in Power System Analysis

Preface

Chapter 6 has a systematic presentation of the Gauss-Seidel method, the exact

is also presented. Chapter 7 has a brief introduction of economic load dispatch and optimal power Chapter 8 deals with the transient stability studies of multi-machine systems with emphasis on numerical simulation. Inclusion of FACTS controllers in stability studies is discussed. Chapter 9 includes a brief discussion of the methods for inclusion of wind

The initial idea a series of two short courses for power engineers at the Memorial University of Newfoundland, Canada. The discussions with Professor H K Kesavan at the University of Waterloo were very helpful in clarifying many of the concepts of graph theory as applied to power systems. The bus impedance algorithm, as developed PhD dissertation of Dr R Jegatheesan at IIT Kanpur, done under the supervision of B Stott, appearing in the July 1974 special issue of Proceeding of IEEE. In the

author wishes to acknowledge gratefully, the support received from IIT Kanpur. For the second edition

support in completing this project. For the third edition, the authors would like to express immense sense of gratitude towards Dr Tony B Nguyen for his help in writing Chapter 7 on Optimal Power Flow. In addition to his contributions towards the content and the examples, including the economic load dispatch part. The authors would also like to thank Mr Arghya Mitra of IIT Kharagpur for his help in the computer solutions of the examples in Chapter 9 (Wind Energy System). M A Pai University of Illinois, USA Dheeman Chatterjee IIT Kharagpur 14 April, 2014 India

McGraw-Hill Education (India) invites suggestions and comments from you, all of which can be sent to [email protected] (kindly mention the title and author name in the subject line). Piracy-related issues may also be reported.

1 The power industry is undergoing vast changes on a global basis. Perhaps the vertically integrated utility (VIU) owning generation, transmission, and distribution

producers (IPPs), will operate in a competitive environment to supply power to the end users. The transmission network, however, is not amenable to restructuring.

determines which generators get access to the load. Hence, it is called open access companies, should not have a material interest in deciding which generators are addition to the transmission companies, at lower voltages, we may have distribution

power, it was recognized early that power industry deregulation would not occur at

1.2

Computer Techniques in Power System Analysis

addition, it has other environmental impacts also. With nuclear power, there is always

augment power generation.

in nature. They vary over the year and also over the day. Hence, there is a challenge

highly complex one.

1.3

systems.

become important issues. Deregulation poses even more challenging problems such problems, which have generally been on an approximate and sometimes heuristic

simulation.

supplying power to a particular area), the problems were complex enough to make pioneering research in synchronous machine theory and transmission lines (which

security assessment, and preventive control, etc. Thus there exists a continued need

the circuit and system theory, control theory, and computer science. A power system

1.4

Computer Techniques in Power System Analysis

mover (hydro or steam) with its governing mechanism. The prime mover is supplied

and shunt reactors, etc. With increased transmission distance, it becomes economic

in achieving this goal without endangering system reliability. modeled either as constant power, voltage dependent, current dependent, or in its

belonging to a particular utility, to be connected to the neighboring utilities through

company may supply any load through the open access transmission network. There no congestion in the network. At the higher level, generators take part in the bidding

Any such decision process must be based on analysis with very heavy emphasis on

1.5

excitation system is the primary device. The power system stabilizer controls the power system oscillations. In addition, some power electronics based controllers have been introduced in the transmission and distribution system. These are connected

or in shunt at a bus to control the reactive power injection and the bus voltage (e.g.

another important issue.

loads on the network is such as to ensure an almost balanced operation on the three

ground, it produces unbalanced currents and voltages in the three phases throughout

1.6

Computer Techniques in Power System Analysis

R, X, and B, parameters. The static capacitors, shown explicitly by their symbols,

the nominal

8 37 25

1

26

29

28

30

38

2 18

27

17

1

24

9

21

35 22

16 6 3 15 10 4

39

14

5 6 7 8

20

13

31 2

11 10

36 33

34 5

9

23

19

12

7

4

32 3

buses. Physically, this is achieved through taps that can be changed in discrete steps automatically or manually depending on voltage deviations at the nearest bus. Tap

real power P and the reactive power Q VI*, where V and I

P

VI* and Q = Imaginary

1.7

a product relationship between the voltage and the current. It thus constitutes a nonlinear component.

can be scheduled at various generating buses so that voltages at the buses are within

interest, P, Q, | V |, and θ which is the angle associated with the phasor V P and Q are V | and θ at the buses given at net P and net Q at all the buses. At each bus, the conservation N on linear N P and Q at all the

P and Q the transmission line losses. In a system, this is generally the generating bus having slack bus or a swing N we relax the assumption that net P and net Q P and | V | and let Q vary within certain limits. This or a P V M N M solution in most cases. In addition to having control over Q at the generator bus, it is

1.8

Computer Techniques in Power System Analysis

the minimum value. This is accomplished by allocating the generation economically minimize the transmission losses. At a higher level, it is possible to do a more sophisticated optimization called the that optimally computes both the real and the reactive power generation taking into account the constraints in the transmission network. In the restructured power system, this is done since it is possible to get prices at any desired node in the network. This will be discussed

between two phases, or a line snapping and making contact with the ground, or a lightning strike hitting a particular transmission line. These give rise to heavy currents

must be isolated. This is done by the circuit breakers actuated through sensing relays.

much greater than the currents during normal operation due to loads, it is customary becomes essentially a linear network problem since the generator will be represented

all the generators run at close to synchronous speeds and the system is said to be in

1.9

I

V

III

II

IV

0.1

1.0

100.0 (seconds)

10.0 log t

Region

Time (in seconds)

I

0 < t < 0.1

II

0.1 < t < 5

III IV V

5 < t < 10 5 < t < 100 t > 60

Phenomena Electromagnetic transient and switching surge Synchronous machine dynamics and its controls Turbine dynamics Load frequency control Boiler dynamics

2 As mentioned in Chapter 1, the power system consists of three main parts, namely (a) generation, (b) transmission, and (c) distribution. Generation involves the synchronous machines along with the exciter, the governor, and the turbine. The transmission system includes two main components: (i) the transformers, and (ii) the transmission lines. The distribution system includes the distribution lines, the transformers, and the different types of loads, namely constant impedance, constant current, and constant power. For simulating/analyzing a power system, models of all these components are required. In this chapter, suitable models of all these components will be discussed. High Voltage DC (HVDC) transmission is being used in power system networks for a long time now. Its advantages over AC transmission for transmitting power over very long distances are well accepted. HVDC links are established in many cases to achieve asynchronous operation of the two sections of the system. With the advancement in power electronics, the HVDC and the system (FACTS) have become integral parts of modern day power systems. Hence, the analysis of the present day power system requires the models of HVDC and FACTS, which are also discussed here. It is not proposed to discuss these in detail, complexity of the model depends to a great extent on the type of study undertaken and the accuracy desired. However, before discussing the modeling, a brief introduction of the per unit system is required. The voltage levels at the generation, the transmission and the such different voltage levels, a per unit system is used which brings the different components of the power system to the same platform. To start with, the method of calculation of the per unit values of different variables is discussed here.

system variables. As the voltage, current, power and impedance of a particular part Usually, the base quantities of the system voltage and the apparent power are chosen

2.2

Computer Techniques in Power System Analysis

Vbase and VAbase respectively. Then, the base for the current (Ibase) and the impedance (zbase) can be calculated as:

I base =

VAbase Vbase

(2.1) 2

Vbase (Vbase ) (2.2) = I base VAbase In case of large power systems, usually the base chosen for the voltage and the power are in kV (kVbase) and MVA (MVAbase) respectively. So in that case, the base for the impedance can be calculated as: zbase =

zbase =

( kVbase )2

(2.3) MVAbase The choice of the base quantity can be totally arbitrary. However, it should be kept in mind that if the phase (line to neutral) voltage is chosen as the voltage base and the per phase power is chosen as the power base, then Eqs. (2.1)–(2.3) can be used. However, if the line to line voltage is chosen as the voltage base and the three-phase

I base =

VAbase 3 Vbase

(2.4)

The zbase can still be found using Eq. (2.3). Usually, the rated voltage and power of the particular power system component is chosen as the kVbase and the MVAbase respectively. The corresponding base values of impedance and current are calculated using Eqs. (2.3) and (2.4). Then, the per unit For example, the per unit impedance is given as: z pu =

zactual (Ω ) MVAbase = zactual (Ω ) zbase ( kVbase )2

(2.5)

In case the base values are changed, the per unit values of the variables in the new base can be computed using the old and the new base values, and the per unit values in the old base as:

( pu )new = ( pu )old ◊

(base)old (base)new

(2.6)

For the analysis of the whole system, one common MVA base is considered for the whole system whereas the kV bases of the respective components are retained. Then, the per unit values of the parameters and variables of the different components with respect to these common bases are obtained using Eq. (2.6). For majority of the studies, the power system is considered to be balanced and be represented by a single phase which is called the single line representation. The single line representation of a 10 machine 39 bus system has been shown in Chapter 1, Fig. 1.1.

Power System Components

2.3

Models of synchronous machines are required for dynamic studies, i.e., the study of the response of the system when it is subjected to some disturbance. Depending on the magnitude of the disturbance, dynamic studies can be small signal (disturbances like small changes in the load) or transient (large disturbances like a fault). In this book, we will not deal with the small signal studies. The detailed model of the synchronous machine is rarely used in transient stability studies. Instead, two different reduced order models of synchronous machines are mostly used, which are: (a) the Flux-decay model, and (b) the Classical model. These representations are valid for dynamic phenomena in the range of 1–5 s. The notations used for the different system variables and parameters while describing the models are as given below. d = the angular position of the rotor with respect to a synchronously rotating reference frame; wr = the rotor speed; ωs = the synchronous speed; H = the inertia constant in sec; kD Pm = the mechanical power in p.u.; Pe = the electrical power in p.u.; xd and xq = the d-axis and the q-axis synchronous reactance respectively in p.u.; x¢d = the d-axis transient reactance in p.u.; E¢q = the q-axis component of the voltage behind the transient reactance in p.u.; T¢do = the d-axis open circuit time constant in p.u.; Id and Iq = the d-axis and the q-axis components of the stator current in p.u.; Efd = the exciter voltage in p.u.; KA and TA = the gain and the time constant of the exciter respectively; V–q = the terminal voltage of the machine, where V is in p.u. and q is in degrees; The terminal voltage of the machine is the same as the voltage of the bus at which the machine is connected with the system. Here all the reactances, voltage magnitudes and the powers are in per unit. E ′q T′do

Flux-decay Model with Fast Exciter The assumptions considered while using this model are listed below. (i) The machine is operating under balanced three-phase positive-sequence conditions. (ii) Synchronous machine damper winding time constants are much less (iii) The stator resistance is neglected. (iv) Turbine-governor dynamics is not considered.

2.4

Computer Techniques in Power System Analysis

jx¢d

V–q

E–f

In Fig. 2.1, the generator internal voltage and its angle, E–f, is given as: ÈÊ ˘ x¢ ˆ E –f = ÍÁ1 - d ˜ V sin (d - q ) + jEq¢ ˙ e j (d -p / 2) ÎË x q ¯ ˚

(2.7)

The machine dynamics are described by the following equations: dd = wr - ws dt k 2H d w r = Pm - Pe - D (w r - w s ) w s dt ws dEq¢

= - Eq¢ - ( xd - xd¢ ) I d + E fd dt The electrical power output, Pe is given as: Tdo¢

Pe = Eq¢ I q - ( xq - xd¢ ) I d I q

(2.8) (2.9)

(2.10)

(2.11)

Neglecting the dynamics, the stator winding is represented by the following algebraic equations: V sin (d - q ) - xq I q = 0

(2.12)

V cos (d - q ) + xd I d - Eq¢ = 0

(2.13)

Replacing the rotor speed (wr) with the per unit speed deviation (Dwr) in Eqs. (2.8) d Dw r 1 dw r and (2.9) using the relations Dw r = (w r - w s ) w s and = , we get dt w s dt dd = w s Dw r dt

(2.14)

Power System Components

2H

d Dw r = Pm - Pe - k D Dw r dt

2.5

(2.15)

Using Eqs. (2.12) – (2.13), Id and Iq are eliminated from Eq. (2.10) and Eq. (2.11) as: Tdo¢

dEq¢ dt

=-

xd Êx ˆ Eq¢ + Á d - 1˜ V cos(d - q ) + E fd Ë xd¢ ¯ xd¢

Pe = Eq¢V sin (d - q ) / xd¢ + 0.5 (1 / xq - 1 / xd¢ )V 2 sin 2 (d - q )

(2.16) (2.17)

If the power output at the generator terminal (Pg + jQg) and the terminal voltage V–θ is known, the internal emf E–φ (classical model) can be obtained as: * Ê P + jQG ˆ I G –g G = Á G Ë V –q ˜¯

(2.18)

E –f = V –q + jxd¢ ( I G –g G )

(2.19)

It can be shown that the rotor angle δ can be obtained as the angle of ÈÎV –q + jxq¢ ( I G –g G )˘˚ . Also it is to be noted that Pg = Pe (as the stator resistance is neglected). (Fig. 2.2). Vref

KA 1 + sTA

+

Efd

V

The dynamics of the exciter can be described as: TA

dE fd dt

= - E fd + (Vref - V ) K A

(2.20)

Classical Model The additional assumptions used are the following: (i) The dynamics of the exciter are not considered in this model. considered as very large, i.e., Tdo¢ = • (iii) Also it is assumed that xq = xd¢ With these assumptions, E¢q becomes constant and equal to its initial value E¢q0. The generator model now consists of a voltage of constant magnitude E¢q0 behind a transient reactance x¢d, as shown in Fig. 2.3. The angle of the voltage is the generator rotor angle d.

2.6

Computer Techniques in Power System Analysis

jx¢d

V–q

E¢q0 –d

The generator equations then become: dd = w s Dw r dt d Dw r = Pm - Pe - k D Dw r dt Where the electrical power output Pe is given as: Eq¢ 0V sin (d - q ) Pe = xd¢ 2H

(2.21) (2.22)

(2.23)

If the power output of the generator at its terminal (Pg + jQg) and the terminal voltage V–q is known, the internal emf E¢q0–d (classical model) can be obtained using Eq. (2.18) and the following equation: Eq¢ 0 –d = V –q + jxd¢ ( I G –g G )

(2.24)

In all the studies to be considered in this book, the transmission network can be assumed to be in sinusoidal steady-state since the time constants associated with it are much smaller compared to the time constants of the synchronous machine.

A transmission line has its resistance, inductance, capacitance, and leakage conductance uniformly distributed along its length and they can be calculated on a per phase, per unit length basis using the parameters of the line. In all practical cases, the leakage conductance to ground is negligibly small. We are essentially interested in the relation between the voltage and the current at the two ends of the line—one of them called the sending end and the other, the receiving end of the line. The distances are measured from the receiving end. There is a progressive change in both the voltage and the current along the line because of the distributed nature of the series impedance and the shunt admittance to neutral. Refer to Fig. 2.4, where, on a per phase to neutral basis, element of length dx at a distance x from the receiving end is concerned.

Power System Components

Is

I + dI

+

I

+

VS

V + dV

2.7

IR +

+

V

VR

dx x=l Sending end

x=0 Receiving end

For this line element, we can write the following relations: dV = Izdx

fi

dV = Iz dx

(2.25)

dI = Vy (2.26) dx where z = series impedance per phase per unit length, and y = shunt admittance per phase per unit length. Differentiation of Eq. (2.1) and Eq. (2.2) with respect to x gives dI = Vydx

fi

d 2V dx

2

d 2I dx

2

=z

dI dx

(2.27)

=y

dV dx

(2.28)

Substituting Eqs. (2.25) and (2.26) in Eqs. (2.27) and (2.28) respectively, we get d 2V dx 2 d 2I dx 2 The solution of Eq. (2.29) is of the form V = A 1 exp

(

= zyV

(2.29)

= zyI

(2.30)

)

(

zy x + A 2 exp - zy x

)

(2.31) x into Eq. (2.25)

results in an expression for the current I as I=

1

(z y)

A 1 exp

(

)

zy x -

1 z y

(

A 2 exp - zy x

)

(2.32)

2.8

Computer Techniques in Power System Analysis

We know that V = VR and I = I R at x = 0 . Using this in Eqs. (2.7) and (2.8), we get A 1=

VR + ( z y ) I R 2

(2.33)

A 2=

VR - ( z y ) I R 2

(2.34)

Here we introduce two new terms—(i) the characteristic impedance of the line Z c = ( z y ) and (ii) the propagation constant g = zy . The expressions of the voltage and the current at a distance x from the receiving end can be written using Eqs. (2.33) and (2.34) in Eqs. (2.31) and (2.32) respectively, as

V ( x) = I ( x) =

VR + I R Z c V -I Z exp (g x ) + R R c exp ( -g x ) 2 2

(VR Z c ) + I R 2

exp (g x ) -

(VR Z c ) - I R 2

exp ( -g x )

(2.35) (2.36)

Equation (2.35) can be rewritten as: 1 1 (exp (g x ) + exp (-g x )) + I R Z c (exp (g x ) - exp (-g x )) 2 2 = VR cosh g x + I R Z c sinh g x (2.37)

V ( x ) = VR

Similarly, we can rewrite the expression for current as I ( x ) = I R cosh g x +

VR sinh g x Zc

(2.38)

The voltage and current at the sending end of the line are obtained by setting x = in Eqs. (2.37) and (2.38) as Vs = VR cosh g + I R Z c sinh g Is =

VR sinh g + I R cosh g Zc

(2.39) (2.40)

Suppose we wish to represent Eqs. (2.39) and (2.40) as an equivalent circuit of the form shown in Fig. 2.5 (called the -equivalent). Then from Fig. 2.5, VS = VR + ( I R + VRYp 2 ) Zp = [1 + Zp Yp 2 ] VR + Zp I R I S = ( I R + VRYp 2 ) + VS Yp 1 = ( I R + VRYp 2 ) + [(1 + Zp Yp 2 )VR + Zp I R ]Yp 1 = [(Yp 1 + Yp 2 ) + Zp Yp 1Yp 2 ]VR + (1 + Zp Yp 1 ) I R

(2.41) (2.42)

Figure 2.5 being the equivalent circuit representation, Eqs. (2.41) and (2.42) should be identical to Eqs. (2.39) and (2.40) respectively.

Power System Components

2.9

For this to happen, we must have Zp IS +

+

VS

Yp1

Yp 2

-

IR

VR -

π

Zp = Z c sinh g

(2.43)

Yp 1 = Yp 2 = Yp

(2.44)

(1+ Zp Yp ) = cosh g

(2.45)

cosh g - 1 1 g tanh = Z c sinh g Zc 2

(2.46)

Hence,

Yp =

Substitution of Zc and g by their respective expressions in Eqs. (2.43) and (2.46), and subsequent rearrangement results in the following: Zp = Z c g

Yp =

(g

sinh g g

=z

sinh g g

(2.47)

2) tanh (g 2) y tanh (g 2) = g 2 Zc 2 g 2

(2.48)

p representation becomes as shown in Fig. 2.6. Zp IS

+

+

IR Zp = z�

VS

Yp

Yp

-

VR

sinh g �

g� y� tan h (g �/2) Yp = g �/2 2

-

π

The terms Zp and Yp can be calculated to any degree of accuracy desired from the tables for hyperbolic functions. In most of the cases, power series expansion with calculation up to two or three terms will give acceptable accuracy.

2.10

Computer Techniques in Power System Analysis

sinh x = x +

x3 x5 + + 3! 5 !

(2.49a)

cosh x = 1 +

x2 x4 + + 2! 4!

(2.49b)

x3 2 5 17 7 + x x + 3 15 315

(2.49c)

tanh x = x -

È (g )2 ˘ Zp ª z Í1 + ˙ 6 ˚ Î Yp ª

y 2

(2.50a)

È 1 Ê g ˆ 2 ˘ y È (g )2 ˘ Í1 ˙ Í1 - ÁË ˜¯ ˙ = 12 ˚ Î 3 2 ˚ 2 Î

(2.50b)

In the case of medium lines (length between 80 km and 240 km), it can be assumed that g 1, Y2 is an inductance and Y3 is a capacitance. For a > 1, Y2 is a capacitance and Y3 is an inductance. As ‘a’ approaches unity, both the shunt approaches the value Yt. Ipq

I¢pq

Yt /a

+ p

q +

Vp

Yt

(1 - a) a2

Yt

(a - 1) a

Vq

- 0

0 -

If the transformer impedance is referred to the tap side (Fig. 2.15), then the -equivalent circuit is given in Fig. 2.16. a:1 Y¢t = Yt/a2

Ipq

I¢pq

p

q

Ipq

I¢pq

aY¢t

+ p

Vp

q + (1 - a)Y¢t

- 0

a(a - 1)Y¢t

Vq 0 -

p

Power System Components

2.15

In interconnected power systems involving loop circuits or parallel lines, the real i.e., a transformer whose turns ratio is a complex number having a certain magnitude and phase angle. The amount of the phase angle depends on the tap position. a–a :1 Ip

Ip

Yt

p

q 1

The derivation of the equivalent circuit for a phase-shifting transformer is somewhat more complex, but we can have a two-port description. The phase shifting transformer can be represented as in Fig. 2.17. The phase angle step may not be

Vp Vr

= a– a = a (cos a + j sin a ) = as + jbs

(2.62)

where α is the phase shift from the bus r to p. It is positive when Vp leads Vr. Since the transformer is ideal, we have, V p I *p = - Vr I q*

(2.63)

Hence, Ip =

-Vr* V p*

Iq = -

1

(as - j bs )

Iq =

Yt (Vr - Vq )

(as - j bs )

(2.64)

Substituting for Vr from Eq. (2.62) Ip =

Vq ˆ Vp Ê Vp Yt Ê ˆ - Vq ˜ = Yt Á 2 Á 2 ¯ (as - j bs ) Ë (as + j bs ) Ë as + bs as - j bs ˜¯

(2.65)

From Eq. (2.64), I q = - ( as - j bs ) Ip

(2.66)

Substituting for Ip from Eq. (2.65), Iq =

-Yt V p

(as + j bs )

+ Yt Vq

(2.67)

2.16

Computer Techniques in Power System Analysis

Combining Eqs. (2.65) and (2.67), we get -Yt ˘ È Yt Í 2 2 ˙ V I a È p˘ as + bs s - j bs ˙ È p ˘ (2.68) Í ˙ Í ˙=Í ˙ ÎVq ˚ Î I q ˚ Í -Yt Yt Í ˙ Î as + j bs ˚ Note that the admittance matrix is not symmetrical and hence we cannot have an equivalent circuit representation. If the turns ratio is real, as = a, b = 0, i.e., there is no phase shift, the equivalent circuit will reduce to that in Fig. 2.12.

Three-winding transformers are used in situations where voltages at two different levels are to be supplied from the primary. The other two windings are called the secondary and the tertiary windings (Fig. 2.18). Tertiary windings are also sometimes used for other purposes such as connecting shunt capacitors or for suppression of third harmonics. The equivalent circuit for the three-winding transformer will now be derived. We shall designate 1-1¢ as the primary winding p, 2-2¢ as the secondary winding s and 3-3¢ as the tertiary winding t. We neglect the exciting or the magnetizing current. p

s

1

2

1¢

2¢

3

t p = Primary winding s = Secondary winding t = Tertiary winding

3¢

The following parameters can be measured experimentally: Zps = Impedance of primary with secondary short circuited and tertiary open. Zpt = Impedance of primary with tertiary short circuited and secondary open. Z¢st = Impedance of secondary with tertiary short circuited and primary open. Suppose Z¢st were referred to the primary side, then, Z st = ( N p N s ) Z st¢ 2

(2.69)

with Zp, Zs, and Zt referred to the primary side. Then, from an interpretation of the short circuit measurements,

Power System Components

2.17

Zps = Zp + Zs

(2.70)

Zpt = Zp + Zt

(2.71)

Zst = Zs + Zt

(2.72)

Zp

Zs

1

2

Zt

3 0

Subtracting Eq. (2.72) from Eq. (2.71), we get Zpt – Zst = Zp – Zs

(2.73)

From Eqs. (2.70) and (2.73), we get Zp =

1 ( Zps + Zpt - Zst 2

)

(2.74)

Zs =

1 ( Z ps + Zst - Zpt 2

)

(2.75)

Zt =

1 ( Zpt + Zst - Z ps ) 2

(2.76)

Similarly,

Ground bus 0 is shown isolated because we are neglecting the shunt impedances of the transformers. Terminal 1 is connected to the supply and terminals 2 and 3 to the load. If the tertiary winding is used for the suppression of harmonics, then terminal

know the variation of the real and reactive power with the variation of voltage. At a typical bus, the load may consist of the following: 1. Induction motors, 50–70% 2. Heating and lighting, 20–30% 3. Synchronous motors, 5–10%

2.18

Computer Techniques in Power System Analysis

Although it would be accurate to consider the P-V and Q-V characteristics of each of these loads for simulation, the analytic treatment would be very complicated. For analytic purposes, there are mainly three ways of representing the load: This is often used for the representation of loads in stability studies when there are convergence problems or when simple models are used for synchronous machines. As the name suggests, the loads are by the load is (PL0 + jQL0), the nominal voltage at the load terminal is V0, and the nominal current drawn by the load is I0, then,

( PL 0 + jQL 0 ) = V0 I 0* fi I 0 =

( PL 0 - jQL 0 ) V0*

(2.77)

2

ZL =

Therefore,

V0 V0 = I 0 ( PL - jQL )

(2.78)

where ZL is the value of the load impedance. As the impedance is considered constant, the load power changes with any change in the voltage at the load terminal. For a voltage value V, the power (PL + jQL) can be computed as: Ê V0 ˆ Ë V0 ˜¯

2

( PL + jQL ) = ( PL 0 + jQL 0 ) Á

(2.79)

In this case, the load current I0 is computed as: I0 =

( PL 0 - jQL 0 ) V0*

= I 0 – (q - f )

(2.80)

ÊQ ˆ where V0 = V0 –q and f = tan -1 Á L 0 ˜ . Here φ is the power factor angle. The Ë PL 0 ¯ magnitude of I0 is held constant. Therefore, the load power varies with the variation of load terminal voltage as: Ê V0 ˆ Ë V0 ˜¯

( PL + jQL ) = ( PL 0 + jQL 0 ) Á

(2.81)

assumed to be constant. That means there is no variation of the power with the load In some cases, the load power at a bus of the power system is considered as a mix of all the three types. Then the total load power is represented as: ˘ È Ê V0 ˆ 2 Ê V0 ˆ a a + + ˙ 3 2 ˜ ÁË V ˜¯ ÍÎ Ë V0 ¯ ˙˚ 0

( PL + jQL ) = ( PL 0 + jQL 0 ) Ía1 Á

Here a1 – a3 are the proportions of each type of load.

(2.82)

Power System Components

2.19

FACTS controllers are mainly of two types, namely the thyristor controlled variable impedance type and the switching converter type. A brief description of each commonly used FACTS controllers are given below.

This type of FACTS controllers use line commutated thyristors as the switching device. A combination of a number of inductors and capacitors are used. Some or all of these circuit elements are brought in or taken out of the circuit (transmission thyristors. This results in a change in the effective reactance of the combination as seen by the transmission system. Thus, a variation in the reactive power exchange between the FACTS controller and the power system is affected. However, this reactive power is proportional to the square of the bus voltage or the line current. Depending on the connection of the controller to the system, they can be (i) shunt type or (ii) series type. The shunt connected variable impedance type FACTS controller is termed as static VAR compensator (SVC). SVC can have different (FC-TCR), and thyristor switched capacitor thyristor controlled reactor (TSC-TCR). Similarly, the are—thyristor switched series capacitor (TSSC), and thyristor controlled series capacitor (TCSC). bus ‘ j’

bus ‘ j’

V–q

V–q i

ic

il

i

xl xsvc

xc T2

(a)

T1

(b)

2.20

Computer Techniques in Power System Analysis

Capacitor voltage, Inductor current

Variation of capacitor voltage inductor current, a = 90 100 Capacitor voltage Inductor current 50

0 - 50

-100 1.48

1.485

1.49 Time(s)

1.495

(a)

Capacitor voltage, Inductor current

Variation of capacitor voltage inductor current, 90 < a < 180 400

Capacitor voltage

300

Inductor current

200 100 0 -100 -200 -300 1.48

1.485

1.49 Time(s) (b)

α 90°

1.495

90° α 180°

Fixed Capacitor-thyristor Controlled Reactor The single-line representation of the FC-TCR is shown in Fig. 2.20(a). The FC-TCR consists of an inductor with with the inductor-thyristor combination. The values of the reactances of the inductor and the capacitor are xl and xc respectively. This combination is connected to the power system bus ‘j’ in shunt, so that the voltage across the capacitor is the phase value of the bus voltage V–θ. The same voltage is applied across the inductor when the thyristor is ON. The anti-parallel thyristors are used for the two half cycles of the supply voltage. The thyristor T1 is switched ON in the +ve half cycle and the current

Power System Components

2.21

T1 remains ON till the current reaches its natural zero. Switching of the thyristor T2 in the –ve half cycle results in the same sequence T1 and T2 are switched ON at a phase lag of 90 from the positive going zero crossing and negative going zero crossing respectively of the voltage across the capacitor (bus voltage in this case), then the inductor current will be continuous. This is clear from Fig. 2.21(a). On the other hand, if the switching is done at a phase lag of 180 , no current will α (90 < α < 180 This angle α can be shown that the amplitude of the fundamental component of this discontinuous current is a function of α. Hence, the effective reactance offered by the thyristor controlled reactor (TCR) denoted as xtcr, is also a function of α. Neglecting the harmonics, the total impedance of the FC-TCR combination, xsvc is given as [24]: xsvc = b1 ( xC + b 2 ) - b 4 b5 - xc

(2.83)

where, b1 =

x x 2 (p - a ) + sin 2 (p - a ) , b 2 = c l , b3 = p xc - xl

b 4 = b3 tan [ b3 (p - a )] - tan (p - a ) , b5 =

xc , xl

4b 22 cos 2 (p - a ) p xl

Thus, the FC-TCR can be equivalently represented by a variable reactance, xsvc as shown in Fig. 2.20(b). This reactance can be inductive or capacitive depending on the value of α. The variation of xsvc with α is shown in Fig. 2.22. Variation of xsvc with firing angle 3

Reactance (p.u.)

2 1 0 -1 -2 -3 100

120 140 160 Firing angle (degree) svc

a

180

2.22

Computer Techniques in Power System Analysis

The values of xc and xl are 0.5 p.u. and 0.1667 p.u. respectively. A positive value of xsvc means the reactance is inductive, whereas a negative value of xsvc means it is capacitive. The minimum value of the capacitive reactance offered is equal to xc, when α is 180 . On the other hand, the minimum inductive reactance offered is equal to the parallel combination of xc and xl, when a is 90 . The reactance becomes a = 130 . Actually at this point, the effective reactance of the TCR, xtcr xc. Usually operation in the very high reactance region (e.g. 120 < a < 140 for this set of values of xc and xl) is avoided due to stability issues. Thyristor Switched Capacitor–Thyristor Controlled Reactor Here, antiparallel thyristors are also provided in series with the capacitor (Fig. 2.23), so that the capacitor can be switched ON or OFF the circuit. However, the switching of the when the instantaneous value of the supply voltage matches the voltage on the capacitor due to its pre-charge. Any mismatch of voltage may lead to huge transients. Therefore, a capacitor will be/ will not be present in the circuit for the complete cycle and hence, the delay angle control of the reactance is not possible for the capacitor. Several capacitors are placed in parallel, with anti-parallel thyristors in series with each capacitor. Depending on the required value of the capacitive reactance (denoted by xtsc), the required number of capacitors are kept in the circuit. Value of the inductive reactance provided by the single TCR (xtcr) is controlled using the delay angle (α) control as in FC-TCR. The net reactance of TSC-TCR can be computed using Eq. (2.83) with xc replaced by xtsc. bus ‘ j’ V–q

icn

T2n

xcn

ic1

T1n

T21

xc1

il

T11

T2

xl

T1

Thyristor Controlled Series Capacitor Single-line representation of the TCSC is shown in Fig. 2.24. Similar to the FC-TCR, it consists of an inductor with a pair inductor-thyristor combination. However, unlike the FC-TCR, this FACTS controller is connected in series with a transmission line, say line i-j between the buses i and j.

Power System Components

2.23

through the capacitor. The zero of the delay angle (α) of the thyristors is counted from the positive going zero crossing and the negative going zero crossing of the voltage across the capacitor. The net reactance offered by the TCSC, xtcsc, is a function of α and can be computed using Eq. (2.83). The equivalent representation of the TCSC, shown in Fig. 2.24, has this variable reactance, xtcsc, in series with the impedance of the line i-j (zline). T2 xl bus ‘ i’

Vi – qi

zline

il

i

ic

bus ‘ j’ T1 Vj – qj xc

bus ‘ i’

bus ‘ j’ zline

Vi – qi

i

xtcsc Vj – qj

This type of FACTS controller is made up of voltage sourced converters (VSC) or current sourced converters (CSC). The source voltage/current is controlled such that the converter exchanges reactive power with the power system. This reactive power support does not depend on the system bus voltage or the line current. Moreover, the use of devices with turn-off capacity increases the speed of response and hence improves the dynamic performance of these controllers. Depending on the connection, switched converter based FACTS controllers can be of (i) shunt type, (ii) series type, (iii) shunt-series type, and (iv) series-series type. The shunt connected controllers are called the static synchronous compensators (STATCOM) and the series connected controllers are called the static synchronous series compensators (SSSC). The shunt-series FACTS controller consists of two converters—one connected in shunt with one of the buses of the power system, and the other connected in series with a transmission line connected to that bus. The converters themselves are connected back to back through a common DC capacitor. controller also has two back to back connected converters sharing a DC bus, but the

2.24

Computer Techniques in Power System Analysis

AC sides of both the converters are connected in series with two different transmission

presence of the back to back connection through the DC bus. A single-line representation of the STATCOM is shown in Fig. 2.25(a). It consists of a DC-AC converter whose AC side is connected to a power system bus ‘m’ through a coupling transformer of leakage reactance jxt. Usually, a capacitor forms the DC side of the converter. However, if the STATCOM has to provide active power support for a short duration, then some energy source (like battery) is connected at the DC side. The current injected by the STATCOM at bus ‘m’, |Ish|–y, can be obtained as: I sh –y =

( Vsh –q sh - Vm –q m ) jxt

(2.84)

where |Vsh|–θsh is the output voltage of the DC to AC converter, |Vm|–θm is the voltage of the bus ‘m’, and xt is the leakage reactance of the coupling transformer. bus ‘ m’

bus ‘ m’

|Vm | –q

|Vm | –qm

|Ish| –y

|Ish| –y Coupling transformer

xt |Vsh| –qsh

Coupling transformer

xt |Vsh| –qsh

DC-AC switching converter +

Variable voltage source representing DC to AC converter

Cs

DC bus capacitor (a)

(b)

Hence, the power injection at bus ‘m’ is given as Psh - jQsh = Vm* I sh = ( Vm – - q m ) ( I sh –y ) Ê V –q sh - Vm –q m ˆ = ( Vm – - q m ) Á sh ˜¯ Ë jxt

(2.85)

Power System Components

Psh =

Vm Vsh sin (q sh - q m ) xt

Qsh =

Vm Vsh V cos (q sh - q m ) - m xt xt

2.25

(2.86) 2

(2.87)

From this, it is clear that the active power exchange will be zero under the condition qsh = qm

(2.88)

It is assumed that the active power loss in the converter and the coupling transformer is zero. Applying Eq. (2.88) to Eq. (2.87), the reactive power exchange between the STATCOM and the bus ‘m’ becomes: Qsh =

Vm ( Vsh - Vm xt

)

(2.89)

The value of |Vsh| determines the direction (and hence nature) of the reactive m’ voltage (|Vm|), then the reactive power is injected into the power system, whereas if the line mid-point voltage magnitude is higher, then reactive power is drawn from the power system. |Vsh| and θsh can be varied by controlling the switching of the converter so that the STATCOM exchanges power as per the requirement of the system. Thus, the converter can be represented by a variable voltage source as shown in Fig. 2.25(b). Similar to STATCOM, SSSC also consists of a DC-AC converter whose AC side is connected to the power system using a coupling transformer. However, unlike STATCOM, the connection is in series with a transmission line m-n (between buses ‘m’ and ‘n’) as shown in the single-line representation, Fig. 2.26(a). The voltages of the two buses are |Vm|–θ and |Vn|–0 . The transmission line reactance is xline, the leakage reactance of the transformer is jxt and the output (ac) voltage of the converter is |Vse|–θse. m-n, |I|–φ, can be obtained as ( Vm –q + Vse –q se - Vn –0∞) (2.90) j ( xt + xline ) The power exchanged by the converter with the power system is given as I –f =

Pse + jQse = Vse I * = ( Vse –q se ) ( I – - f ) Ê V –q + Vse –q se - Vn –0∞ ˆ fi Pse - jQse = Vse* I = ( Vse – - q se ) Á m ˜¯ (2.91) Ë j ( xt + xline ) From which the expressions of the active and the reactive power exchanged are obtained as Pse =

Vse [ V sin (q - q se ) - Vn sin (- q se )] ( xt + xline ) m

(2.92)

2.26

Computer Techniques in Power System Analysis

|Vm| –f

xline

Bus m

| I | –f

|Vn| –0°

Bus n

Coupling transformer

xt |Vse| –qse DC-AC switching converter +

Cs

DC bus capacitor (a) |Vm| – q

|Vse| – qse

xline

xt

| I | –f

bus m

|Vn| – 0°

bus n

(b)

Qse =

Vse [ V cos (q - q se ) - Vn cos (- q se ) + Vse ( xt + xline ) m

]

(2.93)

n’, Pmn + jQmn = Vn I * = ( Vn –0∞) ( I – - f ) Ê V –q + Vse –q se - Vn –0∞ ˆ fi Pmn - jQmn = Vn* I = ( Vn –0∞) Á m ˜¯ Ë j ( xt + xline ) Pmn =

Hence,

Vm Vn V V sin q + se n sin q se ( xt + xline ) ( xt + xline )

(2.94) (2.95)

2

Qmn =

Vm Vn Vn V V cos q + se n cos q se ( xt + xline ) ( xt + xline ) ( xt + xline )

(2.96)

controlled by varying the values of |Vse| and θse, the VSC output voltage magnitude, and its angle. In the process, the power exchanged by the VSC also changes, as is

Power System Components

2.27

evident from Eqs. (2.92) and (2.93). The VSC is represented by a variable voltage source in the equivalent representation shown in Fig. 2.26(b). Usually, the SSSC is not provided with any source of energy and hence does not supply any active power. It is clear from Eq. (2.91) that the SSSC will exchange reactive power only (and no real power) if the voltage injected by it is in quadrature p 2 Using Eq. (2.97) in Eq. (2.90) and subsequent rearrangement, q se = f ±

( I ( xt + xline ) Vse ) –f = [ Vm sin q + j ( Vn - Vm cos q )]

(2.97)

(2.98)

from which we can get, I =

( Vm sin q )2 + ( Vn - Vm cos q )2 ± Vse ( xt + xline )

Ê V - Vm cos q ˆ f = tan -1 Á n Ë Vm sin q ˜¯

(2.100)

Ê V - Vm cos q ˆ p q se = tan -1 Á n ± Ë Vm sin q ˜¯ 2

Hence,

Pmn =

Vm Vn

( xt + xline )

sin q +

Vse Vn

( xt + xline )

(2.99)

(2.101)

p sin ÊÁ f ± ˆ˜ Ë 2¯

(2.102)

2

Qmn =

Vm Vn Vn V V p cos q + se n cos ÊÁ f ± ˆ˜ Ë 2¯ ( xt + xline ) ( xt + xline ) ( xt + xline )

q se = f ±

p = 0 fif = 2

p 2

(2.103)

(2.104)

Using this in Eq. (2.100) gives: Vm sin q = 0, i.e., either Vm = 0 or q = 0

(2.105)

q = 0 means Pmn = 0. Hence none of these is a valid operating condition. Therefore, not possible with the SSSC. A single-line representation of the connection to a power system is shown in Fig. 2.27(a). It consists of one converter connected in shunt and another converter connected in series with a transmission line. Individually, the converters act as variable voltage sources just like the STATCOM and the SSSC. But the converters are connected by a common DC bus capacitor,

2.28

Computer Techniques in Power System Analysis

both the active and the reactive power with the power system. The reactive power is generated/absorbed by the converters. On the other hand, the active power supplied (to the system) by one converter comes from the other converter through the DC bus as the other converter draws an equal amount of active power (from the system). Thus the algebraic sum of the active power exchanged by the two converters is zero (neglecting active power loss of the converters and the coupling transformers). Hence, the angles of the output voltages of the converters are not restricted by the relations in Eqs. (2.88) and (2.97) and can be anything between 0 and 360 . This enables the independently. bus m

bus n xline

|Vm| –qm

|Vn| –q n | I | –f

|Vsh| –y

xt

xt

|Vsh| –qsh

|Vse| –qse DC-AC switching converter

DC-AC switching converter

cs

(a) bus m

|Vse| –qm

bus n xt

xline |Vn| –q n

|Vm| –qm | I | –f

|Ish| –y

xt |Vsh| –qsh

Connection through common DC bus

(b)

variable voltage sources connected in shunt and series with the transmission line, and

Power System Components

2.29

the sources have coupling among themselves. The voltages of these shunt and series sources are |Vsh|–θsh and |Vse|–θse respectively. The current, the active power, the the transmission line, the active and the reactive power exchanged by the series contransmission line can be obtained using Eqs. (2.90), (2.92), (2.93), (2.94) and (2.95) respectively.

A typical HVDC system structure is shown in Fig. 2.28. It consists of converters at both ends, connected to the AC system through the converter transformers. The DC sides of the converters are connected by the HVDC transmission line. However, if the establishment of an asynchronous link is the only objective, then the line may not be there, both the converters may be placed in the same sub-station with their DC sides connected by a link. The HVDC line/link has two conductors—one having positive polarity and the other having negative polarity. This is called bipolar structure. HVDC + Line Converter

AC System

DC System

AC System

The basic converter structure is shown in Fig. 2.29. It consists of six thyristors, T1 – T6. The AC side is represented by a voltage source with per phase source reactance, Ls. Two thyristors are connected to each phase, one for the positive half cycle and the other for the negative half cycle. with positive voltage across the thyristors (Vsw). The phase angular delay between the instant when the Vsw α. The thyristors gets switched OFF automatically when the current through them becomes approximately zero. If the source reactance is neglected (Ls = 0), then the switching ON of the thyristors and hence the change

2.30

Computer Techniques in Power System Analysis

Converter P + Va

Ls

Ia

Vb

Ls

Ib

T1

T3

T5

n

Vd Vc

Ls

Id

Ic

T4

T6

T2 N

in the current through the thyristors takes place instantaneously. This is shown in Fig. 2.30(a), where the phase voltage and the phase/line current of the source is denoted by Vs and Is. The DC side current is assumed ripple free and is represented by a constant current source of value Id. It can be seen from Fig. 2.30(a) that the 1

Vs Is Fundamental of Is

0

-1 0

50

100

150

200

250

300

350

(a) Vs, Is, fundamental of Is for Ls = 0 1 Vs Is Fundamental of Is 0

-1 0

50

100

150

200

250

300

Angle (deg) (b) Vs, Is, fundamental of Is for Ls non-zero

350

Power System Components

2.31

current supplied by the AC source takes the shape of a stepped square waveform, with the peak value being equal to Id. The fundamental of this current is also shown Ls π 0, then the switching cannot take place instantaneously. So the change of current Is takes place over a small time and the current waveform takes the shape of a trapezoid, as shown in Fig. 2.30(b). This period over which the current changes, is called the overlap period and the corresponding angle is called the overlap angle (u extinction angle (d ). Therefore it is clear that δ = a + u. The extinction angle can be found by the following relation: cos d = cos a -

2w Ls 2VL - L

Id

(2.106)

The amplitude of the fundamental of the current is not same for the two cases, but the difference is very small and can be neglected. The r.m.s. value of this fundamental component (Is1) can be found using Fourier analysis as: 2 3 6 I d fi I s1 = I d = 0.78 I d (2.107) p p The average DC voltage available at the converter terminal Vd, is given by 2 I s1 =

Vd = where Vd 0 =

3 2 3 VL - L cos a - w Ls I d = Vd 0 cos a - Rs I d p p

(2.108)

3 2 3 VL - L and Rs = w Ls and VL-L is the AC line-to-line voltage. p p

The AC-side power (Pac) is obtained from the voltage and the fundamental of the current. This must be equal to the DC-side power (Pdc). 3 Pac = 3VL - L I s1 cos f = Pdc = Vd I d = I d ÊÁ Vd 0 cos a - w Ls I d ˆ˜ Ë ¯ p

(2.109)

For Ls = 0, and using Eqs. (2.107) and (2.108), we get: 3VL - L I s1 cos f =

p

I s1.

3 2 VL - L cos a p

6 The reactive power drawn by the converter is:

fi cos f = cos a

Qac = 3VL - L I s1 sin f = 3VL - L I s1 sin a

(2.110)

(2.111)

However, for Ls π cos f =

1 (cos a + cos d ) 2

(2.112)

With Ls = 0, if 90 < a £ 180 , Vd becomes negative. The current direction remaining

2.32

Computer Techniques in Power System Analysis

to the AC side and thus the converter works as an inverter. For Ls π 0, this transition . In the case of the inverter, it is the usual practice to use an angle b instead of a where b b=p–a

(2.113)

Using this in Eq. (2.107), we get the expression for average DC voltage of the inverter as Vd = Vd 0 cos (p - b ) - Rs I d = - (Vd 0 cos b + Rs I d )

(2.114)

The negative sign indicates that the current enters the positive terminal as opposed

by a line of resistance Rl, can be represented by the equivalent circuit shown in can be found as: Id =

Vd 0 r cos a - Vd 0i cos b Rcr + Rl + Rci

Rcr

Rl

(2.115)

Rci

Id Var

Vd0r

Rectifier

Vd0r cos a Vdr

Vdi

Line

Vd0i cos b

Vd0i

Vai

Inverter

power system, such as the transmission lines, the transformers, the loads, and the generators. All these representations are adequate for the basic system studies of FACTS devices, and HVDC transmission system are also presented. We next proceed to interconnect these components into a dynamical interconnected power system and then analyze it. A thermal generating station has two 250 MVA, 22 kV, 50 Hz units. The two units are represented by a single equivalent generator modeled by the Classical model having parameters as: x¢d = 0.4 pu, H = 2 s on the equivalent machine base. The station is connected to the rest of the power system (represented by an

Power System Components

2.33

The transformer is rated at 600 MVA, 22 kV/220 kV with leakage reactance of 10%. The lumped reactances of the 220 kV parallel transmission lines are 82.3 W (line 1) and 118.6 W terminal bus (bus 1) voltage (Vt) is 23.1 kV. The power supplied by the generator at Q = 55.1 MVAR. With 100 MVA as the common (a) Find the phase angle (θt) of the generator terminal bus voltage and the magnitude (in pu) and the angle of the voltage behind the transient reactance of the generator. (b) Write the swing equation of the generator under this condition. Vt – qt

P Q

Line 1

μ Equivalent generator

Infinite bus

Line 2 1

3

2

First, all the system quantities are to be converted to per unit on the common 100 MVA base. Using Eq. (2.3), the base impedance for the transmission lines = zbase =

2202 = 484 W 100

82.3 = 0.17 484

118.6 = 0.245; 484

Equivalent reactance, xeq =

0.17 ¥ 0.245 = 0.10 0.17 + 0.245

Using Eq. (2.6), leakage reactance of the transformer = xt = 0.12 ¥

100 = 0.02 p.u. 600

The generator terminal bus voltage magnitude = 23.1 kV = 23.1/22 pu = 1.05 p.u. The active power transferred = 400 MW = 4.0 p.u. Angle of the generator terminal bus voltage 4.0 ¥ (0.02 + 0.10) ˆ = qt = sin–1 ÊÁ ˜¯ = 27.203° Ë 1.05 ¥ 1.0 I=

1.05–27.203∞ - 1.0–0∞ (4.00 + j 0.55) p.u. j (0.02 + 0.1)

Transient reactance of the equivalent machine = xd¢ = 0.4 ¥

100 = 0.08 p.u. 2 ¥ 250

2.34

Computer Techniques in Power System Analysis

Voltage behind the transient reactance, Eq¢0 –d = 1.05–27.203∞ + j 0.08 ¥ ( 4.0 + j 0.55) = 1.1965–41.96∞ p.u. For writing the swing equation, the inertia constant of the equivalent machine in the system base, H system = H machine ¥

Machine MVA base 500 = 2¥ = 10 System MVA base 100

So the swing equation of the machine becomes: d 2d dt

2

=

180 ¥ 50 È 1.1965 ¥ 1.0 4.0 sin d ˙˘ = 900 ( 4.0 - 5.9825 sin d ) 10 ÍÎ 0.08 + 0.02 + 0.10 ˚

where δ is in degrees. A three-winding 132/33/6.6 kV, three-phase, 50 cycles/s transformer has the following measured impedances between the windings referred to 30 MVA base and 132 kV winding. Zps = 0 + j 0.15 p.u. Zpt = 0 + j 0.09 pu; Zst = 0 + j 0.08 p.u. The 6.6 kV secondary winding supplies a balanced load, taking a current of 2000 A at 0.8 lagging power factor and the 33 kV tertiary supplies a star-connected inductive reactor of (0 + j 132 kV primary terminals to maintain 6.6 kV at the secondary terminals. Zp = j0.08

p

Zs = j0.07

1

2 Zt = j0.01 3 j1.38

0

0

Using Eqs. (2.74) – (2.76) in the equivalent circuit of Fig. 2.33, we get Zs =

1 ( Z ps + Z st - Z pt ) = 12 ( j 0.15 + j 0.08 - j 0.09) = j 0.07 p.u. 2

Zt =

1 1 Z pt + Z st - Z ps ) = ( j 0.09 + j 0.08 - j 0.15) = j.01 p.u. ( 2 2

Zp =

1 ( Z ps + Z pt - Z st ) = 12 ( j 0.15 + j 0.09 - j 0.08) = j 0.08 p.u. 2

Power System Components

Base current =

2.35

Base MVA 30 ¥ 103 Ê 6 ◊ 6 ˆ = = 2624 A Ë 3¯ 3 Base voltage

2000 = 0.76 2624 θ = 0.8. Hence, θ = 36.78°. Taking the secondary voltage as reference = 1– 0°, the current is 0.76 – –36.78° = 0.76 (0.8 – j 0.6) = 0.608 – j 0.456. The voltage at node p is Vp = 1– 0° + j 0.07 (0.608 – j 0.456) = 1.032 + j 0.043 33 kV tertiary winding Load current in p.u.=

Base current =

30 ¥ 103 3 (33)

Base impedance =

= 525 A

Base voltage 33 ¥ 103 = = 36.3 W Base current 3 (525) 50 = 1.38 p.u. 36.3

Current in the tertiary winding =

1.032+j 0.043 j (1.38 + 0.01)

= 0.031 – j0.742 p.u. Current in the primary winding = Current in the secondary winding + Current in the tertiary winding = (0.608 – j 0.456) + (.031 – j 0.742) = 0.639 – j 1.198 p.u. Voltage V1 across the primary winding is given by V1 = (1.032 + j 0.043 + j 0.08 (0.639 – j 1.198)) = 1.13 + j 0.09 = 1.133 – 4.55° p.u.

2.1 Consider a transposed three-phase line whose parameters are z = 0.9 ¥ 10–4 + y = j0.25 ¥ 10–8 j4 ¥ 10–4 Compute (1) the characteristic impedance Zc, (2) the propagation constant , (3) the parameters of the equivalent and T terms of expansion of the hyperbolic terms (Eq. (2.27)). 2.2 receiving end power of 200 MVA at a power factor of 0.8 lagging, compute, 2.3

T circuits. Calculate the sending end voltage and the current for both cases for

2.36

Computer Techniques in Power System Analysis

2.4 Compute the A, B, C, and D parameters for the nominal 2.3, and using these, compute the Y parameters of the line. Verify the Y parameters by writing the node equations at the sending and the receiving ends in the network. From the A, B, C, and D parameters, compute the Z parameters and verify that Z = Y –1. 2.5 A three-winding transformer with primary and secondary rated voltages as 15 kV and 66 kV respectively, has a tertiary winding rated at 5 kV. The measured impedances referred to a 15 MVA base are as follows: Zps = j7.0 p.u., Zpt = j15.0 p.u., and Zst = j10.0 p.u Construct the equivalent circuit. 2.6 and secondary terminals having a series impedance of j6.0 p.u. on a 15 MVA base. Assume a symmetric three-phase to ground short circuit on the 5 kV bus. Compute the fault current. Assume a voltage source of 1 + j0 p.u. connected to the 15 kV bus. 2.7 (a) Consider an off-nominal turns ratio transformer with tap ratio a = 0.8 and an impedance of j 0.05 referred to the unity tap side. Construct the equivalent circuit and identify the shunt elements as capacitive/reactive. (b) Repeat (a) if a = 1.2.

3 The purpose of this chapter is to derive, via a circuit-theoretic viewpoint, the commonly used transmission network representations in power systems. These derivations also provide a logical basis to understand easily the solution of large-scale power networks both for steady state and dynamic analysis. In the new restructured scenario, such a system viewpoint is essential. stability, the transmission network (inclusive of transformers) can be treated as a passive network. The dynamic phenomena in the transmission network are much faster than in the rest of the components so that a steady state representation of the network is reasonable and this fact can be rigorously established via the theory of singular perturbations. The dynamics of the transmission network are relevant in the 0.005 – 0.05 s range and are known as electro-magnetic transients. For deriving the representations of the transmission network, linear graph theory is found to be a very useful tool. The concept of cut-sets, to be discussed, is useful in system stability studies also. There are several texts on linear graph theory.

In the electric transmission network, we are concerned with the interconnection of transmission lines, transformers, and shunt reactors/capacitors that can be modeled in terms of two terminal passive components called elements as discussed in Chapter 2. The points of interconnection are called buses. The graph of a network represents the manner in which the passive elements and the buses are interconnected. Each two terminal element is represented by a line segment called the edge or the branch. The edges represent the interconnections or topology of the network. In the resulting graph, the buses are termed as nodes. Figure 3.1(a) shows a network consisting of nine elements. Its graph is shown be associated with each edge of the graph in which case it is called an oriented or a directed graph (Fig. 3.1(c)). The directions assigned are consistent with the concept of associated reference directions for a two terminal passive element in circuit theory.

3.2

Computer Techniques in Power System Analysis

+1

(3)

3

(1)

2 4

(4)

(2) 5

6 8

7

9

(5) (b)

(a) 1 3

(1)

(3) (2)

4 5

6 8

2

(4) 7 9

(5) (c)

Consider the element in Fig. 3.2(a) which may be a passive element, current, or a voltage source. The associated reference directions are such that a positive current enters the ‘+’ terminal of the voltage reference and leaves the ‘–’ terminal of the voltage reference. The oriented graph is shown in Fig. 3.2(b). If the element is purely passive as in Fig. 3.3(a), and v and i are the voltage and the current phasors, then v = zi or i = yv, where z and y are the complex impedance and the admittance of the element respectively. The reference direction for the branch in the oriented graph is chosen to agree with the current direction (Fig. 3.3 (b)). If the element is a current source (Fig. 3.4(a)), then the positive orientation of the current source is chosen to agree with the reference direction in the graph (Fig. 3.4(b)). Note that terminal (1) has the + sign and terminal (2), the – sign for the voltage across the current source. If the element is a voltage source (Fig. 3.5(b)), the orientation in the graph is chosen such that the arrow in the graph goes from the positive to the negative terminal of the voltage reference. The current, unlike in conventional circuit analysis, goes from the ‘+’ to the ‘–’ terminal. Thus, while for the passive element the orientation of the graph is consistent with the associated reference directions of circuit theory, for the current and the voltage source, it is not.

Topological Analysis of Power Networks

i

(1) +

v

-

(2)

(1)

(a)

i

(1) +

3.3 (2)

(b)

z or y v (a)

-

(2)

(1)

(2) (b)

i (1) +

v

-

(2)

(1)

(a)

i

(1) +

+ v (a)

(2) (b)

-

(2)

(1)

(2) (b)

Consider a network of interconnected components. The passive components may be mutually coupled. The primitive impedance and admittance representation is: v = z i where v and i are vectors, z is the impedance matrix with y as the inverse of z. The diagonal elements of z are self-impedances and the off-diagonal elements are mutual impedances. If the ith and the jth elements are mutually coupled, then the corresponding (i–j and j–i) elements are non-zero. Consider a four terminal network (e.g., three phases of a generator which are mutually coupled) shown in Fig. 3.6, with all unequal mutual impedances.

3.4

Computer Techniques in Power System Analysis

For the passive network, the terminal relations are: Èv1 ˘ È z11 z12 z13 ˘ È i1 ˘ Ív ˙ = Í z ˙Í ˙ Í 2 ˙ Í 21 z22 z23 ˙ Íi2 ˙ ÍÎv3 ˙˚ ÍÎ z31 z32 z33 ˙˚ ÍÎi3 ˙˚ v=zi

v1 v2 +

(1) i1

z1

(2) i2

z12

(3) i3

z23

v3

(3.1) (3.2)

(1) z13 (2)

(0)

z3

-

(3) (0) (a)

(1)

(1)

+-

(2)

(2)

+-

(3)

+-

(3)

e0

(1)

e0 e0

(2)

(0)

e0 (0)

-+

(3)

(0) (b)

Suppose an identical voltage source e0 is introduced between the node (0) and the common terminal with the polarity shown in Fig. 3.6(a). Then it is equivalent to moving the voltage source in series with all the three coils (Fig. 3.6(b)). The graph will remain the same with each edge of the graph representing the Thevenin source. The terminal relations are now È1˘ Í v = z i + 1˙ e0 Í˙ ÍÎ1˙˚ The admittance formulations for Eqs. (3.2) and (3.3) are respectively, i = yv È1˘ Í i = y v - y 1˙ e0 Í˙ ÎÍ1˚˙

(3.3)

(3.4) (3.5)

Topological Analysis of Power Networks

3.5

loads. Hence, only the positive sequence network is considered. The impedances in the per-phase equivalent are known as positive (2) sequence impedances. The calculation of these 4 1 positive sequence impedances for a transmission line (both series impedance and shunt admittance) 3

(1)

in power system analysis. Topologically the positive sequence network is the same as the original singleline diagram of the network. Consider the graph of a certain passive network shown in Fig. 3.7. The primitive impedance v – i relationship is given by Èv1 ˘ È z11 Ív ˙ Í0 Í 2˙ Í Ív3 ˙ = Í0 Í ˙ Í Ív4 ˙ Í0 ÍÎv ˙˚ ÍÎ0 5

0

0

0

z22 0

0

0

z33 0

0

0

z44

0

0

0

2

(4)

5 (3)

0 ˘ Èi1 ˘ 0 ˙˙ ÍÍi2 ˙˙ 0 ˙ Íi3 ˙ ˙Í ˙ 0 ˙ Íi4 ˙ z55 ˙˚ ÍÎi5 ˙˚

(3.6)

The primitive admittance i – v relationship is: i = yv where y = z–1 -1 È z11 Í Í0 Í y = Í0 Í Í0 Í0 Î

0

0

0

-1 z22

0

0

0

-1 z33

0

0

0

-1 z44

0

0

0

0 ˘ ˙ 0 ˙ ˙ 0 ˙ ˙ 0 ˙ -1 ˙ z55 ˚

(3.7)

Graph theory is a vast mathematical discipline with applications in various engineering graph. It is said to be connected edges of the graph is called a subgraph. Certain degenerate subsets such as an edge graph in Fig. 3.7 are shown in Fig. 3.8. The number of edges incident at a node gives the degree of the node. In Fig. 3.7, all other nodes of degree two in the subgraph is called a path

3.6

Computer Techniques in Power System Analysis

edge at the most once. For example, the subgraphs shown in Fig. 3.9 form the paths between nodes (1) and (4). The direction of the path that is arbitrarily drawn for each path is independent of the orientation of its edges. (2)

(2)

1

1 3

(1)

(4)

3

(1) 2

5 (3) (3)

(3)

(a)

(b)

(c)

(2)

(2)

1 3

(1)

4

(4)

3

(1)

5

(4)

2 (3)

(3) (2)

(1)

(4) 1

2

4

5 (3)

(1)

(4)

loop (circuit) is a connected subgraph with the degree of each of the nodes in the is also referred to as a closed path. For Fig. 3.1(c), some of the loops are (1, 2, 4, 3),

the elements of the loop. For the graph in Fig. 3.7, two of the loops are shown in Fig. 3.11 (a) and (b) along with their orientations. Figure 3.11(c) is not a loop since the degree of node (2) in that subgraph is three.

3.7

Topological Analysis of Power Networks (3)

1

2

(1) (2)

3

(4)

4

6

5

6

1

(1)

(2)

3

(1)

(5)

(3)

2

(1) 3

(4)

(2)

8

5

(5)

7

(5)

(2)

(2)

1

(2) 4

1

(1)

(1)

3

(4) 2

2

4

1 3

(1)

5

(3)

5

(3) (b)

(a)

(c)

tree is a subgraph that is connected, contains all nodes, and has no loops. For example, in Fig. 3.1, a are shown in Fig. 3.12. In a tree, there is exactly one path between any two nodes. If the number of nodes in a graph is n, there are exactly (n – 1) edges in a tree. The proof of this observation is obvious. The elements of the tree are called tree-branches. (2)

(2) 1

(1)

(4) (1)

3

1 (4) (1)

5

2

(2) 4

(4)

3

5

5

(3)

(3)

(3)

(a)

(b)

(c)

Those edges of the graph that are not in a tree form a co-tree, and the edges of the co-tree are called links or chords

3.8

Computer Techniques in Power System Analysis

is a co-tree. For the three trees chosen in Fig. 3.12, the corresponding co-trees are shown in Fig. 3.13. In general, a co-tree does not contain all the nodes of the graph of several subgraphs (Fig. 3.13(c)). Figure 3.14 shows another example of a graph, a tree, and the corresponding co-tree. If the total number of edges in a graph is e, then the number of links = e – (n – 1) = e – n + 1. (2)

(2)

(2) 4

4

1 (1)

(4)

3

(1)

(1)

(4) 2

2

(3)

(3) (b)

(a)

(1)

(1)

(4)

(4)

4

(2)

(3)

(0)

2

5

3

3 6

(4)

4

2

(2)

(1)

1

1 5

(c)

(3) (2)

6

(3)

(5)

(5)

(a)

(b)

(c)

fundamental loop for a graph is formed from the tree of the graph by inserting an appropriate link. For each link inserted, we create a new fundamental loop in the tree. There will be in all (e – n + 1) fundamental loops for a chosen tree, all of these being linearly independent. These are linearly independent because each fundamental loop contains a new link. For the graph of Fig. 3.1(c), repeated in Fig. 3.15(a), let the tree chosen be (1, 4, 8, 9) as in Fig. 3.15(b) (solid lines). 1 (1) 3 6

1

2

(3)

(4) (1)

(2)

3

4 5

8

6

7 9

(5)

2

(3)

(4)

(2)

4 5

8

7 9

(5)

Topological Analysis of Power Networks

3.9

By inserting the links 2, 3, 5, 6, 7 one at a time, the following e – n + 1 = 9 – 5 + 1 = 5 fundamental loops are generated. (The links are underlined.) (1, 2, 4, 9, 8), (3, 4, 9, 8), (5, 9, 4), (6, 8), (9, 7)

Fundamental Loop Matrix, through the application of Kirchhoff’s voltage law (KVL). KVL states that for any closed path or loop, the algebraic sum of voltages around the loop is zero. 1. Select a tree. 2. For each fundamental loop, assign a positive reference direction to agree with the orientation associated with the link for that loop. 3. Going around the loop along the reference direction, assign a + sign to the voltage of the edge if the orientation of the edge agrees with the reference direction, a – sign if it is the opposite, and a zero if the edge is not contained in that loop. 4. Repeat step 3 for all the fundamental loops. the link voltages afterwards. 6. The resulting matrix of +1, –1, and 0 entries is called the Fundamental Loop Matrix. Consider the graph of Fig. 3.16(a) and the tree (1, 3, 4) in Fig. 3.16(b). Draw the corresponding fundamental loops and hence write the KVL in matrix form. (1)

1

(2)

2

(3)

(1)

(2)

1

3 5

(1)

1

(3)

3 4

4 (4)

(4)

(a)

(b)

(2)

2

(3)

(1)

1

3

(2)

(3)

3 5

4

4

(4)

(4)

(c)

(d)

,

,

, and

3.10

Computer Techniques in Power System Analysis

The fundamental loops obtained by inserting links 2 and 5 are shown in Fig. 3.16 (c) and (d). The positive reference direction for each fundamental loop is for the two loops are written as v2 + v3 – v4 = 0 v1 + v3 – v5 = 0

[v1 v3 v4

v2 v5

(3.8)

]T.

The KVL equations can now be put in a matrix form as Tree-branches Links È v1 ˘ (3.9) 1 3 4 2 5 Í v3 ˙ Í ˙ 2 È0 1 -1 1 0˘ Ív4 ˙ 5 ÍÎ1 1 0 0 1 ˙˚ ÍÍv ˙˙ 2 Ív ˙ Î 5˚ For Fig. 3.1(c), choose the tree (1, 2, 4, 5) and write the KVL in matrix form. The tree is shown in Fig. 3.17 in solid lines and the links in dotted lines. 1

2

(3)

(4)

(1) 3

(2)

4 5

6

7

8

9 (5)

The KVL equations can be written by inspection as

1 2 Ê -1 -1 Á -1 -1 Á Á 0 0 Á -1 -1 Á Ë 0 0

4 1 1 1 1 1

5 0 -1 -1 -1 -1

3 1 0 0 0 0

6 0 1 0 0 0

7 0 0 1 0 0

8 0 0 0 1 0

9 0ˆ 0˜ ˜ 0˜ 0˜ ˜ 1¯

Èv1 ˘ Ív ˙ Í 2˙ Ív4 ˙ Í ˙ Ív5 ˙ Ív ˙ = 0 Í 3˙ Ív6 ˙ Í ˙ Ív7 ˙ Ív ˙ Í 8˙ ÍÎv9 ˙˚

(3.10)

Topological Analysis of Power Networks

3.11

Generalization equations can be written in the form È n -1 e - n +1˘ È v ˘ e - n + 1 ÍCb : U ˙ Í- b ˙ = 0 Î ˚ Î v ˚

(3.11)

i.e., C.v = 0 where Cb is a (e – n + 1) ¥ (n – 1) matrix. U is a (n – 1) square matrix. vb is a subvector of order n – 1 corresponding to the tree-branch variables. v is a subvector of order e – n + 1 corresponding to the link variables. C is called the fundamental loop matrix. The existence of the unity submatrix in C (i) Each fundamental loop contains one link only, and (ii) The positive orientation of the loop coincides with the orientation of the link for that particular loop. In general, the entries of C are such that 1. Cij = + 1 if the element corresponding to the j th column is in the fundamental i th row and their orientations agree. 2. Cij = – 1 if the element corresponding to the j th column is in the fundamental i th row but their orientations are opposite. 3. Cij = 0 if the element corresponding to the j th column is not in the i th row. For the transmission network shown in Fig. 3.18(a), assume that the shunt admittances at each bus are lumped into a single admittance. The oriented graph is shown in Fig. 3.18(b) with (0) representing the ground bus. Pick a tree and write the fundamental loop matrix C. (1)

1

(1)

(2)

(2)

3

2

4

5 (3)

(4) 9

6 8

7 (3)

(4) (a)

(0) (b)

The following tree is chosen with the tree-branches (2, 4, 7, 8), shown by solid lines. The links are: (1, 3, 5, 6, 9) shown by dotted lines. The orientations for the fundamental loops are shown dotted.

3.12

Computer Techniques in Power System Analysis

C tree-branches 2

links

4

7

8

1 3 5

1 Ê -1 -1 3 Á 0 -1 Á C =5 Á 0 0 6 Á -1 0 Á 9 Ë 0 1

-1 -1

1 1

-1

1

1 0 0 0 0ˆ 0 1 0 0 0˜ ˜ 0 0 1 0 0˜ 0 0 0 1 0˜ ˜ 0 0 0 0 1¯

-1 0 0 -1

6

9 (3.12)

exactly two parts. Consider the graph in Fig. 3.19(a). Removal of the elements (3, 4, 5, 6, 7) (Fig. 3.19(b)) leaves the graph in three parts as shown in Fig. 3.19(c). Note that node (5) by itself constitutes a subgraph. Hence, (3, 4, 5, 6, 7) does not form a cutset. On the other hand, removal of (4, 6, 7) (Fig. 3.19(d)) leaves it in (2)

(1)

10

7

2

6 5

(6)

(7)

9

5

2

8

1 (4)

11

10

6

8

(3)

(2)

(7)

9

11

(1)

7

(5)

4

4

(3) 1

(6)

3

3

(4) (a)

(c)

(b) (1)

(2)

7

10 (7)

9

11

6

8

(3)

2

(6) 5 1

4 (4) (d)

(5)

3 (e)

Topological Analysis of Power Networks

3.13

two parts as shown in Fig. 3.19(e). Hence (4, 6, 7) is a cutset. The elements of the cutset can also be selected by “cutting” the graph with a curved (dotted) line not passing through any node and dividing the graph in two connected subgraphs. It can also be noted that the nodes of the graph also get divided in two groups by the cutset (4, 6, 7)—one group consisting of nodes (1), (3), (4), and (5), and the other group consisting of nodes (2), (6), and (7). The edges of the cutset connect the nodes between the two groups as shown in Fig. 3.20. The reader may verify the other cutsets in Fig. 3.19(a) as (2, 11, 7), (1, 2, 3, 4), (4, 6, 9, 10), (2, 4, 6, 11), etc. Just as the concept of fundamental loops is associated with a link, so is the concept of fundamental cutsets associated with a tree-branch that we discuss next. (1)

7

(3)

6

(4) (5)

(2) (6) 4 (7)

The tree is a connected subgraph of a given graph. Removal of any tree-branch two groups of nodes. The edges of the graph connecting these two groups of nodes are called fundamental cutsets and they correspond to that particular tree branch. The edges of the cutset are the particular tree-branch and other links that connect the two groups of nodes. Thus, for each tree-branch we have an associated fundamental cutset. In all, we have (n – 1) fundamental cutsets since a tree in an n node graph has (n – 1) edges. Consider the graph in Fig. 3.21(a). Let the tree-branches be (2, 4, 5, 7) which constitutes a connected graph (Fig. 3.21(b)). Removal of tree-branch 2 in the tree possible links of the graph between the two nodes. This constitutes a fundamental cutset associated with branch 2. For convenience, tree-branch 2 is shown in a solid line and the other links are shown dotted. The fundamental cutsets corresponding to the other tree-branches, 4, 5, and 7 are similarly shown in Figs. 3.21(d), (e), and (f) respectively. To avoid this laborious procedure, we can follow the simple rule of cutting the graph by a curve not crossing any node such that it cuts only one treebranch at a time. This is shown in Fig. 3.21(g). Thus the fundamental cutsets for the graph in Fig. 3.21(a) and the chosen tree in Fig. 3.21(b) are (2, 1, 6), (4, 1, 3), (5, 1, 3), and (7, 6). (The tree-branches are underlined.) It is of interest to remark here that a set of linearly independent cutsets can also Fig. 3.22. The elements incident on each node is a cutset, and the edges of each cutset are the ones cut by a curved line. But we shall see later that only (n –1) cutsets in an n node graph constitute a linearly independent set of cutsets.

3.14

Computer Techniques in Power System Analysis

(5)

(5) 6 (2) 1

(2)

(3)

2

(1)

7

7

3 4

2

(3) (2)

2

5

5 (1)

(4)

(1)

3 4

(5)

(4) 4 (c)

(2)

1

(4)

6

(b)

(a)

(1)

1

(3)

(3)

(1)

(4)

(4)

(2)

1 3

(3) 5 (5)

(5) (d)

(e)

(5)

(1) 6 (5) 7

7

6

(2)

(2)

(3)

1

(3) 2 3

5

(4) (1)

(f)

Just as in the case of fundamental loops, we shall use KCL to derive another important topological

4

(4)

(g)

(1)

(2)

(4)

(3)

(Fig. 3.16) as for KVL, and it is reproduced in Fig. 3.23. Choose (1, 3, 4) as the tree. The three fundamental cutsets associated with tree-branches are shown by the dotted curved lines. These are (1, 5), (2, 3, 5), and (2, 4). The underlined element corresponds to the tree-branch. If corresponding to each fundamental cutset, the curved dotted line were extended to form a closed surface, then KCL states that the algebraic sum of the currents leaving a closed surface is zero. To apply KCL

Topological Analysis of Power Networks

3.15

give a + sign to an edge of the cutset if its orientation agrees with the orientation of the cutset and a – sign if it is the opposite. The application of KCL to each of the three cutsets in Fig. 3.23 gives 1

(1)

2

(2)

(3)

3 5

4

(4)

i1 - i5 = 0; -i2 + i3 - i5 = 0; i2 + i4 = 0 Tree-branches 1 3 4 1 È1 0 0 Tree-branches 3 Í0 1 0 Í 4 ÎÍ0 0 1

Links 2 5 È i1 ˘ 0 -1˘ Íi3 ˙ = 0 Í ˙ -1 -1˙ Íi4 ˙ ˙ 1 0˙˚ Íi ˙ Í 2˙ Íi ˙ Î 5˚

(3.13)

(3.14)

For the graph in Fig. 3.24 and tree (1, 2, 4, 5), write the KCL.

(3)

1 3

(1)

2 4

(4)

(2) 5 8

6

7 9

(5)

The fundamental cutsets are shown in Fig. 3.24 along with their positive orientations shown by an arrow in the direction coinciding with that of the tree branch.

3.16

Computer Techniques in Power System Analysis

The KCL is written as Tree-branches 1 2 4 5

Links 3 6 7

È1 Í0 Í Í0 Í Î0

0˘ 1 1 0 1 0˙ ˙ -1 -1 -1 -11 -1˙ ˙ 0 1 1 1 1˚

0 0 0 1 0 0 1 0 0

0 0 1

1

1

0

8 1

9

Èi1 ˘ Íi ˙ Í 2˙ = 0 Íi4 ˙ Í ˙ Íi5 ˙ Íi ˙ Í 3˙ Íi6 ˙ Í ˙ Íi7 ˙ Íi ˙ Í8˙ ÍÎi9 ˙˚

(3.15)

Generalization For a general graph, we can write the KCL as È n -1 n - 1Í U Î

e - n +1

˘ Èib ˘ B ˙ Í ˙=0 (3.16) ˚ Îi ˚ Since each fundamental cutset contains only one tree-branch, the nature of the unity matrix U is self-evident. In a more compact form, where

Bi = 0 B=[U|B ]

(3.17) (3.18)

and i is the vector of currents arranged in the order of the tree branch and the link currents. B is called the fundamental cutset matrix. It has a unity submatrix of order (n – 1) in the leading position, and the matrix B of order (n – 1) ¥ (e – n + 1) in the B are such that bij = 1 if the orientation of the element corresponding to the j th column agrees with the orientation of the tree-branch corresponding to the i th row. bij = –1 if the orientation of the element in the j th column is opposite to the orientation of the tree-branch (2) 1 (1) corresponding to the i th row. bij = 0 if the element corresponding to the j th column does not belong to the 4 2 3 tree-branch corresponding to the i th 9 6 row. (4) (3) 5

For the graph in Fig. 3.18(b) and the chosen tree (2, 4, 7, 8), write the B matrix. The graph is redrawn in Fig. fundamental cutsets.

7

8 (0)

Topological Analysis of Power Networks

3.17

The B matrix is written by inspection as tree branches 2 4 7 8 2 Ê1 4 Á0 B= Á 7 Á0 8 ÁË 0

1

3

links 5 6

9

1 0ˆ -1 -1 0 0 1˜ ˜ 1 1 1 1 0˜ -1 -1 -1 0 1˜¯

0 0 0

1

1 0 0 0 1 0 0 0 1

0

0

(3.19)

One of the characterizations of a graph is the incidence matrix. The edges incident to a node in a graph is called the incidence set. Thus a connected graph has as many + sign to the currents leaving a node and a – sign to the currents entering the node. node and the positive orientation of the cutset outwards from the dotted closed line (see Fig. 3.26). (1)

(2)

1

2

(3)

3

5

4

(4)

The KCL equations for nodes (1) – (4) can be written as i1 - i5 = 0 - i1 - i2 + i3 = 0

(3.20)

i2 + i4 = 0 - i3 - i4 + i5 = 0 In matrix form, these can be written as 1 (1) (2) Nodes (3) (4)

2

Edges 3 4

5

È i1 ˘ Ê 1 0 0 0 -1ˆ Í ˙ i Á -1 -1 1 0 0˜ Í 2 ˙ ˜ Íi3 ˙ = 0 Á Á 0 1 0 1 0˜ Í ˙ ÁË 0 0 -1 -1 1˜¯ Íi4 ˙ ÍÎi5 ˙˚

(3.21)

3.18

Computer Techniques in Power System Analysis

In a more compact form, Aa i = 0 In general, the order of Aa is n ¥ e where n = number of nodes and e = number of edges in the graph. Aa is called the node to branch incidence matrix or the augmented incidence matrix. The entries of Aa are such that (aij)a = +1 if the edge corresponding to the j th column is incident to the node corresponding to the i th row and is directed away from it. (aij)a = –1 if the edge corresponding to the j th column is incident to the node corresponding to the ith row and is directed towards the node. (aij)a = 0 if the edge corresponding to the j th column is not incident to the node corresponding to the i th row. It may be observed that since each element is incident on two nodes, the columns of the Aa matrix have each a +1 and a –1 entry. If we add up all the rows of Aa matrix, we get a zero row. This indicates that the rows are linearly dependent. The number of linearly independent rows are n – 1, or we say that the rank of the matrix Aa is (n and the resulting matrix A is called the reduced-incidence or simply the incidence matrix. The order A is (n – 1) ¥ e. In power networks, if a ground bus is present, it is generally taken as the reference bus and the node corresponding to it is generally deleted while writing the A matrix. If the network has no connection to the ground, any one of the nodes is taken as the reference.

in Fig. 3.18. Choose the ground bus (0) as the reference bus. By inspection, the matrix A is written as

A = Nodes

(1) (2) (3) (4)

Edges 1 2 3 4 5 Ê 1 1 0 0 0 Á -1 0 -1 -1 0 Á Á 0 -1 1 0 1 ÁË 0 0 0 -1 -1

6 7 8 9 1 0 0 0ˆ 0 0 0 1˜ ˜ 0 1 0 0˜ 0 0 1 0˜¯

A B C, In the matrix A, the columns corresponding to the edges were arranged sequentially. They can be written in any particular order. In fact, one of the ways is to arrange the columns in the order of tree branches and links for a given tree in the graph. Thus, we can write A as Tree branches Links A = [ Ab

A

]

(3.22)

Topological Analysis of Power Networks

3.19

The following properties are now stated without a rigorous proof and illustrated with some examples. The proofs can be found in texts on graph theory.

Property 1 For a given tree of a graph, each row of the fundamental loop matrix C is orthogonal to each row of the fundamental cutset matrix B. Mathematically, this relationship implies BCT = CTB = 0

(3.23)

Since B = [U | B ] and C = [Cb | U], it follows

[U

B

È C bT ˘ ˙=0 ÎU ˚

]Í

(3.24)

Therefore, CbT = - B which is the same as C b = - BT

(3.25)

This is a very important result. It tells us that for a given tree of a graph, if the fundamental loop matrix C is known, the fundamental cutset matrix is also known

Property 2 Let the incidence matrix A be arranged in the order of tree-branches and links for a given tree, i.e., n -1 e - n +1 A = n - 1 ÈÎ Ab

A

(3.26)

˘˚

It can be shown that Ab is nonsingular. Furthermore, the fundamental cutset matrix for the given tree is given by B = Ab-1 A = Ab-1 [ Ab | Al ] = ÈÎU | Ab-1 Al ˘˚

(3.27)

since B = [U | Bl ] , \ Bl = Ab-1 Al This important result tells us that by choosing a tree and writing the incidence matrix by inspection (or computer generated), we can obtain the fundamental cutset matrix B and from property 1, also the fundamental loop matrix C. Consider the graph shown in Fig. 3.27. Choose the tree whose branches are (1, 3, 5). Find the fundamental cutset and loop matrices B and C using the incidence matrix A. Choosing (2) as the reference node, we write the reduced incidence matrix A as

(1)

1

(4)

5

2 (2) 3

4 (3)

3.20

Computer Techniques in Power System Analysis

Tree branches

Links

Nodes 1 3 5 (1) Ê 1 0 0 A = (3) Á 0 -1 0 Á (4) Ë -1 0 -1 = [Ab

2 4 1 0ˆ 0 1˜ ˜ 0 -1¯

A ]

-1 È 1 0 0˘ È 1 0 0˘ Í ˙ = 0 -1 0 = Í 0 -1 0˙ Í ˙ Í ˙ ÍÎ-1 0 -1˙˚ ÍÎ-1 0 -1˙˚

Ab-1

(3.28)

(3.29)

Therefore, Ab-1

Hence,

È 1 0 0˘ A = Í 0 -1 0˙ Í ˙ ÍÎ-1 0 -1˙˚ È 1 0˘ = Í 0 -1˙ Í ˙ ÍÎ-1 1˙˚ Tree branches

1 3 5 1 Ê1 0 0 B = 3 Á0 1 0 Á 5 Ë0 0 1 = [U B

È 1 0˘ Í0 1˙ Í ˙ ÍÎ0 -1˙˚

(3.30)

Links 2 4 1 0ˆ 0 -1˜ ˜ -1 1¯

(3.31)

]

Since Cb = - BT , we have Tree branches 1 3 5 C=

2 Ê -1 0 1 4 ÁË 0 1 -1

links 2 4

(3.32)

1 0ˆ 0 1˜¯

The nature of matrices B and C

In electric power networks, the most useful representations are the multiport characterizations of the transmission network. The network consists of the devices (capacitors or inductors), and the loads. The terminals are the n buses plus the additional ground bus, i.e., (n circuit, or transient stability study, one is interested in the terminal quantities only,

Topological Analysis of Power Networks

3.21

namely the voltages of the buses with respect to ground, and the current entering the buses (injected currents). Thus the ground bus is implicitly assumed as a common n terminal pairs or n ports as shown in Fig. 3.28. I1 +

(1) (2)

+

I2

V1

V2

(k)

+

Ik

Vk In

- -

+ Vn - -

(n) (0)

Let the currents and voltages be phasors so that the voltage and current vectors denoted by Vbus and Ibus respectively, are È I1 ˘ ÈV1 ˘ Í ˙ Vbus = , I bus = Í ˙ (3.33) Í ˙ Í ˙ ÍÎVn ˙˚ ÍÎ I n ˙˚ The relationship between Vbus and Ibus can be characterized either in an impedance or an admittance form as Ibus = Ybus Vbus

(3.34)

Vbus = Zbus Ibus

(3.35)

-1 Z bus = Ybus

(3.36)

where

Zbus is called the open-circuit impedance matrix in circuit theory with the elements of the matrix known as the driving point and transfer impedances. The reason for this designation will be explained shortly. In power literature, it is known as the bus impedance matrix. Ybus is called the short-circuit admittance matrix, or nodal admittance matrix, with the elements of the matrix known as the driving point and transfer admittances. In power system literature, it is known as the bus admittance matrix.

assume that the graph of the four nodes is connected.

3.22

Computer Techniques in Power System Analysis

The three-port description can be explicitly written as ÈV1 ˘ È Z11 ÍV ˙ = Í Z Í 2 ˙ Í 21 ÎÍV3 ˚˙ ÍÎ Z 31 i.e.,

Z13 ˘ Z 23 ˙ ˙ Z 33 ˙˚

Z12 Z 22 Z 32

I1

(1)

+ I2

È I1 ˘ ÍI ˙ Í 2˙ ÍÎ I 3 ˙˚

(2)

+ V1 I3

V2

Vbus = Zbus Ibus - -

(3)

+ V3 -

(0)

network, the element Zbus can be obtained by external measurements as follows. 1. Excite terminal (1) with a current source I1 = 1.0 – 0° with the current entering terminal (1) and leaving the reference bus. The other terminals are left open-circuited. 2. Compute or measure V1, V2, V3 which will be numerically equal to Z11, Z12, and Z31 Zbus . 3. Excite terminal (2) with a current source I2 = 1.0 – 0° with the current entering into terminal (2) and leaving the reference bus. The other terminals are open. Compute or measure V1, V2, and V3 to get the second column of Zbus, namely Z12, Z22, and Z32. 4. Repeat step 3 with terminal (3) to get the third column of Zbus. Thus, we get all the impedance parameters. Determine the Zbus of the p-section of the transmission line shown in Fig. 3.30. z1

I1 (1)

(2) I2

+

I1

+

(1)

+ I2 V1

z3

z2

V2

-

V1 -

-

(2)

+ V2 -

(0)

(a)

(b) z1

I1

I2

+ 1.0

V1

+ z2

z3

(c)

Zbus

V2

Topological Analysis of Power Networks

3.23

Excite terminal (1) with a unit current as shown in Fig. 3.30 leaving

V1 = Z11 =

z2 ( z1 + z3 ) z1 + z2 + z3

(3.37)

V2 = Z 21 =

z2 z3 z1 + z2 + z3

(3.38)

Similarly, by exciting terminal (2) and leaving terminal (1) open-circuited, we can get Z12 =

z2 z3 z (z + z ) ; Z 22 = 3 1 2 ; z1 + z2 + z3 z1 + z2 + z3

Therefore, Z bus =

z2 z3 ˘ 1 È z2 ( z1 + z3 ) ; Í z3 ( z1 + z2 )˙˚ z1 + z2 + z3 Î z2 z3

(3.39)

For the four-terminal network in Fig. 3.29, the admittance representation will be I1 = Y11V1 + Y12V2 + Y13V3 I2 = Y21V1 + Y22V2 + Y23V3 I3 = Y31V1 + Y32V2 + Y33V3

(3.40)

Yij’s can be computed as follows from the external measurements. 1. Excite terminal (1) with voltage source = 1.0 – 0° with the positive terminal of the source at bus (1). The terminals (2) and (3) are short-circuited to the ground (0). 2. Compute or measure I1, I2, and I3 which will be numerically equal to Y11, Y21, and Y31 Ybus. 3. Repeat steps 1 and 2 at terminals (2) and (3) to get the second and third columns, respectively. Suppose the p-section parameters of Fig. 3.30 are given as admittances as in Fig. 3.31. Express the elements of the Ybus matrix in terms of these admittances. Excite terminal (1) with a voltage source equal to unity and short-circuit terminal (2) as shown in Fig. 3.31(c). It is easily seen that I1 = Y11 = y1 + y2,

I2 = Y21 = –y1

(3.41)

Similarly, exciting terminal (2) with a unit voltage source and short-circuiting terminal (1), we get Y12 = – y1;

Y22 = y1 + y3;.

3.24

Computer Techniques in Power System Analysis

È y1 + y2 Ybus = Í Î - y1

Thus (1)

I1

+ 1 V1 y2 = z 2

- y1 ˘ y1 + y3 ˙˚

I2

(3.42)

(2) I1

+

1 y1 = z1

(1)

+ y3 =

I2

1 z3 V2

-

V1 -

-

(2)

+ V2 -

(a)

(0) (b)

y1

I1

1 – 0° + -

y2

y3

I2

(c)

From KCL, the fundamental cutset equations are Bi=0 i.e., Thus

[U

B

Èi ˘

] Í ib ˙ = 0 Î ˚

ib = -B i

(3.43) (3.44) (3.45)

Since B = - CbT (from Eq. (3.25)), so ib = CbT i i = i to Eq. (3.45), we get È ib ˘ È CbT Íi ˙ = Í U Î ˚ Î

˘ T ˙i = C i ˚

(3.46)

CT is known as the connection matrix since it expresses all the currents in terms of the link currents.

Topological Analysis of Power Networks

3.25

Hence the link transformation is i = C Ti

(3.47)

where i represents currents in all the elements of the network. Note the ordering of elements in the vector i. The KVL corresponding to the fundamental loops can be written as

[C b

Èvb ˘ U] Í ˙ = 0 Îv ˚

–Cbvb = v

(3.48) (3.49)

Since Cb = - BT , we have from Eq. (3.49), v = B T vb

(3.50)

vb = vb, we have È vb ˘ È U Í ˙=Í T Î v ˚ ÍÎ B i.e.,

˘ ˙ vb ˙˚

v = BTvb

(3.51) (3.52)

Thus in Eq. (3.52), the voltages in all the elements of the network are expressed in terms of the tree-branch voltages. This is known as the tree-branch transformation. V. Hence, the fundamental cutset and loop matrices contain important information regarding the dependence of link voltages on tree-branch voltages, and tree-branch currents on link currents, respectively. To be v = B T vb

(3.53)

ib = CbT i

(3.54)

The nodal transformation to be discussed next exploits the incidence matrix A discussed earlier. Consider an n bus network with an additional ground bus (0) so that there are (n + 1) buses in total. Let the bus voltages be measured with respect to the ground bus. The voltages of each branch of the network may be expressed in terms of the bus voltages as follows. For the kth branch, vk = Vi – Vj if in the graph, the edge k is directed from node (i) to ( j). = Vj – Vi if in the graph, the edge k is directed from node ( j) to (i). = ± Vi if node (i) is connected to the ground bus, with a + sign if it is directed from node (i) towards the ground bus, and a – sign if it is directed from the ground bus to node (i).

3.26

Computer Techniques in Power System Analysis

In matrix form, we can write it as

È v k = Í0 Í ÍÎ

or

or

È v k = Í0 Î

j

i

-1

1

i

-1

j

1

ÈV1 ˘ Í ˙ Í ˙ ˘ ÍVi ˙ Í ˙ 0˙ Í ˙ ˙ ˙˚ ÍV j ˙ Í ˙ Í ˙ ÍV ˙ Î n˚

(3.55)

ÈV1 ˘ Í ˙ Í ˙ ÍVi ˙ ˘Í ˙ 0˙ Í ˙ ˚ ÍV ˙ j Í ˙ Í ˙ ÍV ˙ Î n˚

(3.56)

ÈV1 ˘ ÍV ˙ Í 2˙ i È ˘Í ˙ vk = Í0… ±1… 0˙ Í ˙ Î ˚ ÍVi ˙ Í ˙ Í ˙ ÎVn ˚

(3.57)

For the e elements in the network, we have È v1 ˘ ÈV1 ˘ e n nodes Í ˙ = edges È1, - 1, or 0 entry ˘ Í ÍÎ ˙˚ Í Í ˙ ÍÎve ˙˚ ÍÎVn ˚

(3.58)

It is easy to see that the e ¥ n matrix is precisely the transpose of the incidence matrix A, i.e.,

v = AT Vbus

(3.59)

This is the nodal transformation. then at the nodes, we can write KCL as Ai = Ibus where I is the vector of currents in the network of dimension n. current graph is

(3.60)

Topological Analysis of Power Networks

V¢bus = –Vbus

3.27

(3.61)

Now from the conservation of power law, * v T i * + (Vbus =0 ¢ ) I bus T

(3.62)

Using Eqs. (3.60) and (3.61), we get * v T i * = (Vbus ) I bus = (Vbus ) Ai * T

T

(3.63)

Since this is true for all the values of i, we have v = AT Vbus

I1 V1

(3.64)

(1)

(1)

(2)

(2)

(n)

(n)

(0)

(0)

I2 V2 In Vn

(a)

(b)

graph in Fig. 3.16(a) for the tree (1, 3, 4). Choose (4) as the reference node. C, B, and A matrices in Eqs. (3.9), (3.14), and (3.21) respectively. A is obtained from Aa by deleting the row corresponding to the Link Transformation i = CTi È i1 ˘ È 0 Íi ˙ Í Í 3˙ Í 1 Íi4 ˙ = Í-1 Í ˙ Í Íi2 ˙ Í 1 Íi ˙ ÍÎ 0 Î 5˚

1˘ 1˙ ˙ Èi2 ˘ 0˙ Í ˙ ˙ Îi5 ˚ 0˙ 1˙˚

Branch Transformation V = BT Vb

(3.65)

(3.66)

3.28

Computer Techniques in Power System Analysis

È v1 ˘ È 1 0 Ív ˙ Í 0 1 Í 3˙ Í Ív4 ˙ = Í 0 0 Í ˙ Í Ív2 ˙ Í 0 -1 ÍÎv5 ˙˚ ÍÎ-1 -1

0˘ 0˙ ˙ 1˙ ˙ 1˙ 0˙˚

ÈV1 ˘ ÍV ˙ Í 3˙ ÎÍV4 ˚˙

(3.67)

By arranging the rows and columns of Aa in Eq. (3.21) and omitting row (4), the incidence matrix A for this graph with (4) as the reference node, is given by Edges 1 3 4 2 5 (1) Ê 1 0 0 0 -1ˆ A= Nodes (2) Á -1 1 0 -1 0˜ ˜ Á (3) Ë 0 0 1 1 0¯

(3.68)

Hence the nodal transformation is given by V = ATVbus È v1 ˘ È 1 -1 Ív ˙ Í 0 1 Í 3˙ Í Ív4 ˙ = Í 0 0 Í ˙ Í Ív2 ˙ Í 0 -1 ÍÎv5 ˙˚ ÍÎ-1 0

0˘ 0˙ ÈV1 ˘ ˙ 1˙ ÍV2 ˙ ˙Í ˙ 1˙ ÍÎV3 ˙˚ 0˙˚

(3.69)

Note that the ordering of the edges in Eq. (3.68) is different from Eq. (3.21). However, the ordering of the voltages in Eq. (3.69) is consistent with the ordering of currents in Eq. (3.66).

Y Consider the n-port network shown in Fig. 3.32(a) excited by current sources. Omitting the graph of the network, the additional edges to the graph corresponding to the current sources are shown in Fig. 3.32(b). The currents are denoted by the vector Ibus and noting the reference directions, the voltages associated with these edges are denoted by the vector –Vbus nodes can be written in matrix form as explained in the previous section as Ai – Ibus = 0

(3.70)

Ibus = Ai

(3.71)

i.e., T

Since v = A Vbus and i = y v, where y is the primitive admittance matrix, we have Ibus = A yATVbus = YbusVbus

(3.72)

Topological Analysis of Power Networks

3.29

Thus the topological formula for Ybus is Ybus = A yAT

(3.73)

Consider the network in Fig. 3.33(a) where two branches have mutual coupling as shown. Find the primitive impedance matrices z, y, and the Ybus matrix. Choose (0) as reference bus. j1

(1)

(2)

j1

1

(1)

(2)

2

j2 (3)

(3) 5

j1

j1

3

4

j1 (0)

(0) (a)

(b)

Ybus

The graph is shown in Fig. 3.33(b). The primitive impedance matrix z is 1 Ê1 Á1 Á z = j Á0 Á0 Á Ë0

2 3 4 5 1 0 0 0ˆ 2 0 0 0˜ ˜ 0 1 0 0˜ 0 0 1 0˜ ˜ 0 0 0 1¯

(3.74)

The primitive admittance matrix y is calculated as z –1. 1 2 3 4 5 Ê 2 1 0 0 0ˆ Á 1 -1 0 0 0˜ Á ˜ y = j Á 0 0 -1 0 0˜ Á 0 0 0 -1 0˜ Á ˜ Ë 0 0 0 0 -1¯

(3.75)

3.30

Computer Techniques in Power System Analysis

The incidence matrix obtained by taking (0) as the reference node is 1

2

3 4

5

(1) Ê 1 1 0 0 1ˆ A = (2) Á -1 0 1 0 0˜ ˜ Á (3) Ë 0 -1 0 1 0¯

(3.76)

AT is given by (1) 1Ê1 2Á1 Á T A = 3 Á0 4 Á0 Á 5Ë1

(2) (3) -1 0ˆ 0 -1˜ ˜ 1 0˜ 0 1˜ ˜ 0 0¯

(3.77)

Hence, Ybus = A y AT È-2 1 0 ˘ = j Í 1 -3 1 ˙ Í ˙ ÍÎ 0 1 -2˙˚

(3.78)

Consider the same network as in Fig. 3.33(a) but with no mutual coupling. Denote the network in terms of the admittances y10, y21, y13, y20, y30 where y is the admittance to ground. The primitive admittance matrix y is 1 Ê y11 Á 0 Á y =Á 0 Á 0 Á Ë 0

2

3

4

5

0

0

0

y12

0 y30

0

0

y40

0

0

0ˆ 0˜ ˜ 0˜ 0˜ ˜ y50 ¯

0 0 0

0

(3.79)

The matrix A remains the same. Performing the triple matrix product, we get the bus admittance matrix Ybus = A yAT (1) (1) Ê y12 + y13 + y10 = (2) Á - y12 Á Ë (3) - y13

(2) - y12 y12 + y20 0

(3)

Y ÈY - y13 ˆ Í 11 12 ˜ = ÍY21 Y22 0 ˜ ÍÎY31 Y32 y13 + y30 ¯

Y13 ˘ Y23 ˙ ˙ Y33 ˚˙

(3.80)

Topological Analysis of Power Networks

3.31

From this pattern, we can generalize and say that when there is no mutual coupling, 1. The (i – j) element Yij of the Ybus matrix is the negative of the sum of admittances connected between nodes (i) and (j), and is called the transfer admittance. 2. The diagonal (i – i) element is the sum of all admittances connected to node (i). It is called the self-admittance. Thus, Ybus can be written by inspection in this case and also easily programmed on the computer. In network theory literature, Ybus is called the Nodal Admittance matrix and is obtained by writing the node equations at all the buses except the ground bus. For the transmission network in Fig. 3.34, the lines 2-3 and 2-4 have half line charging admittances of 0.005 mho and 0.01 mho, respectively. Ybus matrix. The shunt capacitor at bus (4) has an admittance of 0.02 mho. (1)

(1)

(2) j0.2

j0.25

(4)

j0.5

j0.2

(a)

- j4.0

j0.4

(4) (3)

j0 .01

(2 )

- j5.0

j0.005

- j2.0

j0.01

- j2.5

(3)

- j 5.0

j0.005

j0.02

(b)

To make matters clearer, we draw the network diagram for Fig. 3.34(a) with all the element values in terms of admittances as shown in Fig. 3.34(b). The Ybus matrix is given by

Ybus

(1) (2) (3) (4) (1) Ê - j 9 j5 0 j4 ˆ (2) Á j 5 - j 9.485 j 2.5 j2 ˜ ˜ Á = (3) Á 0 j 2.5 - j 7.495 j5 ˜ (4) ÁË j 4 j2 j5 - j 10.97˜¯

The reader can verify the entries Y22, Y33 and Y44 by running all the admittances at nodes (2), (3), and (4) respectively. The off-diagonal entries are the negative of admittances connected between the two nodes.

3.32

Computer Techniques in Power System Analysis

Z Earlier in Section 3.6.1, it was mentioned that the elements Zij of Z can be interpreted as the open-circuit impedance parameters, and computed by exciting the buses with current sources one at a time. To derive an analytical expression for Zij, we shall generalize this procedure by exciting all the buses with current sources simultaneously, and solving for the bus voltages in terms of these current sources. Let the graph of the transmission network consisting of (n + 1) buses, including the ground bus, be augmented by elements corresponding to the current sources connected between the buses and the ground bus, with positive bus current orientation into the buses. Hence, Ibus = [I1, I2, ..., In,]T. In accordance with our convention of the associated reference directions for the elements, the voltages of these augmented edges will be negative of the bus voltages. If V¢bus is the vector of the voltages of these elements, then Vbus = – V¢bus techniques. The overall network now has n + 1 nodes and e + n elements, of which e = number of elements of the network, and n = number of elements corresponding to the current sources. First we choose a tree for the overall augmented network. It will have n treebranches. The tree is chosen such that the current sources constituting the vector Ibus belong to the co-tree. This choice of assigning the current sources to the co-tree is a logical one. From the link transformations, all currents are expressible in terms of the link currents, and hence this choice decreases the number of unknown link currents in the augmented network. The fundamental loop equations for the overall network are written as CU = 0

È e ÍCb Î n

È vb ˘ ˙ Í ˘ …˙ =0 U˙ Í ˚Ív ˙ ˙ Í ¢ ˚ ÎVbus e

(3.81)

In Eq. (3.87), vb, v , and V¢bus are of the order n, e – n and n respectively, with the elements corresponding to vb and v belonging to the transmission network, and the elements of Vbus corresponding to the voltages across the augmented current sources. Expanding Eq. (3.81), n

e-n

n

È vb ˘ 0˘ Í U È …˙ ˙=0 e ÍCb … … …˙ Í Í ˙Ív ˙ ÍÎ 0 U ˙˚ Í ˙ ¢ ˚ ÎVbus

(3.82)

Cb is further partitioned as n ÈCb11 ˘ e - n Cb = Í ˙ ÎCb 22 ˚ n

(3.83)

Topological Analysis of Power Networks

n

n

3.33

e-n

e-nÈ 0 ˘ È Cb11 = -Í Í ˙ ¢ ˚ n ÎVbus ÎCb22

U˘ 0 ˙˚

È vb ˘ Ív ˙ Î ˚

(3.84)

The primitive network equations are arranged in the usual order of tree-branches and links of the transmission network as È vb ˘ È ib ˘ Í ˙ v = Í ˙ = [z] Í ˙ Í ˙ ÍÎv ˙˚ ÍÎ i1 ˙˚

(3.85)

Substituting Eq. (3.85) in Eq. (3.84) and replacing V¢bus = –Vbus, we get È 0 ˘ È Cb11 Í ˙=Í Í ˙ Í ÍÎVbus ˙˚ ÍÎCb 22

U ˘ È ib ˘ ˙ [z]Í ˙ ˙ Í ˙ 0 ˙˚ ÍÎ i ˙˚

(3.86)

From the link transformation Eq. (3.47), we have È ib ˘ Í i ˙ = ÈC T ˘ È i ˘ Í ˙ Î ˚ ÍÎ I bus ˙˚ ÍÎ I bus ˙˚

(3.87)

Using the same partitioning of C as in Eq. (3.84), we get

Hence,

T T È ib ˘ ÈCb11 Cb 22 ˘ ˙È i ˘ Í i ˙=Í U 0 ˙Í ˙ Í Í ˙ Î I bus ˚ ÍÎ I bus ˙˚ ÍÎ 0 U ˙˚

(3.88)

È ib ˘ ÈCbT11 CbT22 ˘ Èi ˘ ˙Í ˙ Íi ˙ = Í 0 ˚ Î I bus ˚ Î ˚ Î U

(3.89)

Substituting Eq. (3.89) in Eq. (3.86), we get ÈCbT11 CbT22 ˘ Èi ˘ È 0 ˘ ÈC b11 U ˘ ˙ [z] Í ˙Í ˙ ÍV ˙ = Í 0 ˚ Î I bus ˚ Î bus ˚ ÍÎC b22 0 ˙˚ Î U

(3.90)

The triple matrix product in Eq. (3.90) is then performed to get È 0 ˘ È Z11 ÍV ˙ = Í Z Î bus ˚ Î 21

Z12 ˘ Èi ˘ Z 22 ˙˚ ÍÎ I bus ˙˚

(3.91)

Solving for i -1 i = - Z11 Z12 I bus

(3.92)

3.34

Computer Techniques in Power System Analysis

Eq. (3.92) is then substituted in the second set of equations in Eq. (3.91). -1 Vbus = - Z 21 Z11 Z12 I bus + Z 22 I bus

(3.93)

Hence,

(

)

-1 Vbus = Z 22 - Z 21 Z11 Z12 I bus = Z bus I bus

(3.94)

Zbus is a square matrix and has the same order as i , i.e.,, e – n which is the number of links in the transmission network alone. This is also equal to the number of fundamental loops in the transmission network, i.e., network without the augmented current sources. Hence, to compute Zbus, we need an inverse of the same order as the number of links in the network, i.e., e – n. In the power network, this number could be very large since there are a number of shunt paths to the ground due to the transmission lines. Therefore, for a large network, the computation of Zbus needs an inverse of a very high order. The Zbus building algorithm avoids this problem by progressively building the Zbus by taking one element at a time. This is a specialized topic and is omitted. Zbus can also be computed from Ybus inversion schemes such as the LU decomposition technique. Hence, with these fast numerical techniques, building Zbus or inverting Ybus has comparable computing requirements. Since Ybus is generally sparse and we may not require all the elements of Zbus such as in short-circuit studies, the LU decomposition technique has some 1 advantages. But in small systems, it is nice to have a program for building Zbus as each element is added or deleted. 2 5

Consider the network of Fig. 3.33. Its graph augmented with the current source edges (6, 7, 8) is shown in Fig. 3.35. Find the Zbus considering the tree with tree-branches (2, 3, 4). The links are (1, 5, 6, 7, 8). The fundamental loop matrix for the augmented graph is 2 1 Ê -1 5 Á -1 Á C =6Á 1 7Á 0 Á 8Ë 0

3

4

1 -1 0 -1 0 1 1 0 0 1

3 6

8

4

7

Zbus 1 5

6 7

8

1 0 0 0 0

0 0 1 0 0

0ˆ 0˜ ˜ 0˜ 0˜ ˜ 1¯

0 1 0 0 0

0 0 0 1 0

(3.95)

Hence, Cb11

È1 0 1 ˘ È-1 1 -1˘ Í ˙ =Í ˙ , Cb 22 = Í0 1 0˙ Î-1 0 -1˚ ÍÎ0 0 1 ˙˚

(3.96)

Topological Analysis of Power Networks

3.35

The primitive impedance matrix is given by (note that the ordering of elements has to be in the same order as the tree-branches and links of the network graph) 2 3 4 1 5 2 Ê 2 0 0 1 0ˆ 3 Á 0 1 0 0 0˜ Á ˜ z = j 4 Á 0 0 1 0 0˜ 1 Á 1 0 0 1 0˜ Á ˜ 5 Ë 0 0 0 0 1¯ The triple matrix product corresponding to Eq. (3.90) is given by

2 1 Ê -1 5 Á -1 Á È 0 ˘ ÍV ˙ = 6 Á 1 Î bus ˚ 7Á 0 Á 8Ë 0

3 4 1 -1 0 -1 0 1 1 0 0 1

1 5 1 5 1 0ˆ 1 -1 Ê Á ˜ 0 1 1 0 Á ˜ 0 0˜ [ z ] Á -1 -1 Á 1 0 0 0˜ Á ˜ Ë 0 1 0 0¯

(3.97)

6 7 8 1 0 0ˆ 0 1 0˜ ˜ È i ˘ (3.98) 1 0 1˜ Í ˙ Î I bus ˚ 0 0 0˜ ˜ 0 0 0¯

Performing the triple matrix product, we get È 3 2 -2 1 -1˘ Í 2 4 -3 0 -1˙ Í ˙ È Z11 È 0 ˘ ÍV ˙ = j Í-2 -3 3 0 1˙ = Í Z Í ˙ Î 21 Î bus ˚ Í 1 0 0 1 0˙ ÍÎ -1 -1 1 0 1˙˚

Z12 ˘ È i ˘ Z 22 ˙˚ ÍÎ I bus ˙˚

(3.99)

Hence, performing the elimination of [i ] in Eq. (3.99)

(

)

-1 Vbus = Z 22 - Z 21 Z11 Z12 I bus = Z bus I bus

Hence,

È3 0 1 ˘ Z bus = j Í0 1 0˙ Í ˙ ÍÎ1 0 1 ˙˚ È3 0 1 ˘ = j Í0 1 0 ˙ Í ˙ ÍÎ1 0 1 ˙˚

È-2 -3˘ -1 È 3 2˘ j Í 1 0˙ Í Í ˙ Î2 4˙˚ ÍÎ -1 -1˙˚

(3.100)

È-2 1 -1˘ Í -3 0 -1˙ Î ˚

È-2 -3˘ 1 È 4 -2˘ È -2 1 -1˘ j Í 1 0˙ Í Í ˙ 8 Î-2 3˙˚ ÍÎ -3 0 -1˙˚ ÍÎ -1 -1˙˚

È3 0 1 ˘ È 2.375 - 0.25 0.875˘ È0.625 0.25 0.125˘ = j Í0 1 0˙ - Í- 0.25 0.5 - 0.25˙ = j Í 0.25 0.5 0.25˙ (3.101) Í ˙ Í ˙ Í ˙ ÍÎ1 0 1 ˙˚ ÍÎ 0.875 - 0.2 ÍÎ0.125 0.25 0.625˙˚ 25 0.375˙˚ By using Ybus Ybus Zbus = U

(3.102)

3.36

Computer Techniques in Power System Analysis

Obtain the Zbus for the transmission network of Example 3.13 (Fig. 3.34). In Fig. 3.34(b), we lump the shunt admittances at a bus into a single admittance, and then draw the graph of the network. The augmented current sources are shown dotted (9, 10, 11, 12) in Fig. 3.36. Choose the tree (1, 2, 4, 6). The treebranches are hatched. The links are loop matrix for the augmented graph. (1)

1

2

(2) 10

3

4

9

8

(3) 12

5

(4)

11

6

7

Zbus 1 2 4 6 3 Ê -1 1 0 0 5 Á -1 1 -1 0 Á 7 Á 1 -1 0 -1 8 Á 1 -1 0 -1 C= Á 9 Á 0 1 0 1 Á 10 Á -1 1 0 1 11 Á -1 1 1 -1 1 Á Ë 12 0 0 0 1

3 5 7 8 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Hence,

Cb11

È-1 1 0 0˘ Í-1 1 -1 0˙ ˙ =Í Í 1 -1 0 -1˙ Í ˙ Î 1 -1 0 -1˚

Cb22

È 0 Í-1 =Í Í-1 Í Î 0

1 0 1 0 1 -1 0 0

1˘ 1˙ ˙ 1˙ ˙ 1˚

9 10 11 12 0 0 0 0ˆ 0 0 0 0˜ ˜ 0 0 0 0˜ 0 0 0 0˜ ˜ 1 0 0 0˜ ˜ 0 1 0 0˜ 0 0 1 0˜ ˜ 0 0 0 1¯

Topological Analysis of Power Networks

3.37

The primitive impedance matrix written in an order consistent with the tree-branches and links of the network is 1

2

4

6

3

5

7

8

0 0 0 0 0 0 ˆ Ê 0.2 0 Á 0 .25 0 0 0 0 0 0 ˜ ˜ Á 0 .04 0 0 0 0 0 ˜ Á 0 Á 0 0 0 -33.333 0 0 0 0 ˜ ˜ z=Á 0 0 0 0.5 0 0 0 ˜ Á 0 ˜ Á 0 0 0 0 0.2 0 0 ˜ Á 0 Á 0 0 0 0 0 0 -200 0 ˜ ˜ Á Ë 0 0 0 0 0 0 0 -66.67¯ The rest of the details of deriving Zbus is left to the reader. The reader may also verify that Zbus Ybus = U.

In large-scale systems, it may be necessary to reduce the dimension of the network by eliminating certain buses. Let the description be of the form Ibus = Ybus Vbus

(3.103)

Ibus and Vbus are ordered such that the buses to be r). Those to be eliminated are denoted by the subscript (s). Thus È I bus(r) ˘ ÈVbus(r) ˘ I bus = Í ˙ and Vbus = Í ˙ Î I bus(s) ˚ ÎVbus(s) ˚

(3.104)

Ibus(r) = YA Vbus(r) + YB Vbus(s)

(3.105)

Partition Ybus such that Ibus(s) = YC Vbus(r) + YD Vbus(s) (3.106) If there are no sources at the buses to be eliminated, then Ibus (s) ∫ 0. Hence, by elementary matrix manipulation from Eqs. (3.105) and (3.106), we get, I bus(r) = ÈÎYA - YB YD-1 YC ˘˚ Vbus(r) = Vbus(r) Vbus(r) (3.107) Instead of taking the inverse of the matrix YD, we can perform the elimination bus by k in Ybus, then, 1 k n

Ybus

Ê Y11 Á Y21 Á Á =Á Y Á k1 Á ÁË Yn1

Y1k Y2k Ykk Ynk

Y1n ˆ Y2n ˜ ˜ ˜ Ykn ˜ ˜ ˜ ˜ Ynn ¯

(3.108)

3.38

Computer Techniques in Power System Analysis

The row and column corresponding to the bus k order the rows and columns so that the k th row and the k th column become the last row and the last column, respectively, resulting in Eq. (3.109). Hence,

Ybus

ÈY11 ÍY Í 21 Í =Í ÍYn1 Í Í ÎYk1

Y1k ˘ Y2k ˙ ˙ ˙ ˙ Ynk ˙ ˙ ˙ Ykk ˚

Y1n Y2n Ynn Ykn

(3.109)

Eliminating the last row and the last column in Eq. (3.110) using (3.107), we get, Yij ( new ) = Yij (old ) -

Yik (old ) Ykj (old ) Ykk

(3.110)

This can be repeated for each node to be eliminated. Kron reduction of this type is useful in transient stability studies. For the network in Fig. 3.33(a), eliminate node (2) and obtain the reduced 2 ¥ 2 description at the nodes (1) and (3).

Ybus

1

Ybus

1 2 3 1 0ˆ 1 Ê -2 = j Á 1 -3 1˜ 2 ˜ Á Ë 0 1 -2¯ 3 2 3

1 3 2 1ˆ 1 Ê -2 0 1 Ê -2 1 0ˆ Á 0 -2 = j Á fi j 1˜ 3 ˜ Á 2 Ë 1 -3 1˜¯ Ë 1 1 -3¯ 2

Eliminating node 2 gives the elements of the reduced Ybus matrix as Y11 new = -2 -

1 5 (1)(1) = -2 + = 3 3 (-3)

Y12 new = 0 -

(1)(1) 1 = (-3) 3

Y21 new = 0 -

(1)(1) 1 = (-3) 3

Y22 new = -8 -

5 (1)(1) =3 (-3)

Topological Analysis of Power Networks

3.39

Hence, the reduced matrix is 1

Ybus

3

1ˆ Ê- 5 Á 3 3˜ 1 Á ˜ =Á ˜ Á 1 5˜ 3 - ˜¯ ÁË 3 3

In Eq. (3.106), if Ibus(s) π 0, then we can still eliminate those buses as follows. From Eq. (3.106), Vbus(s) = YD-1 I bus(s) - YD-1 YC Vbus(r)

(3.111)

Substituting this in Eq. (3.105) yields I bus(r) = ÈÎYA - YB YD-1YC ˘˚ Vbus(r) +YB YD-1 I bus(s) fi I bus(r) - YB YD-1 I bus(s) = ÎÈYA - YB YD-1YC ˘˚ Vbus(r) = Ybus(r) Vbus(r) The elements of the vector -YB YD-1 I bus(s) which are eliminated on the buses which are retained. The elements of the matrix -YB YD-1 are called the distribution factors Dij .

In this chapter we have discussed the systematic way in which the multiport have also discussed Kron reduction and distribution factors.

3.1 For the transmission network in Fig. 3.18(a) and its corresponding oriented graph in Fig. 3.18(b), select the tree T (6, 7, 8, 9) and write the B and C matrices. Verify the orthogonality relations. Choosing ground as the reference bus, write the A matrix. 3.2 For the graph in Fig. P.3.1, select the tree T (2, 4, 5, 6). C and the fundamental cutset matrix T B. Verify the relation BC = 0 and Cb = - BT . Aa and the incidence matrix A by A as [ Ab A ] corresponding to the tree T (2, 4, 5, 6) and verify the relation B = - Ab-1 A .

3.40

Computer Techniques in Power System Analysis

1 1

4

2 2

3

2

4

1

3

4 7

3

3 1

5

2 5

8 5

6 5

4

6 7

6

3.3 Repeat Problem 3.2 for the tree T (3, 5, 6, 7). 3.4 For the oriented graph in Fig. P.3.2, select the tree T B A with 6 as and C matrices and verify that BCT the reference node and derive B from A. 3.5 Consider the linear graph in Fig. P.3.3 which represents a 3-bus transmission system with all the shunt admittances at a bus lumped together. Each transmission line has a series impedance of 0.02 + j 0.08 and a half line charging admittance of j 0.02 (all in p.u.). 0 is the ground bus. (a) Compute Ybus by inspection as well as by the analytical formula (3.73). (b) Compute Zbus analytically by the procedure of Section 3.7.4. (c) Verify Zbus Ybus = U. 3 1

2 0.4

1

2 3

2

1

3 0.4

0.23 5

4

0.6

0.5 0.5

0.3 0

0.2

6

4

5

6

3.6 Consider the system shown in Fig. P.3.4. It shows a transmission network with series reactances of transmission lines all in p.u. as shown. The line charging and shunt impedances are neglected. (a) By choosing the appropriate tree for the graph, write the B and C A with 6 as the reference node. (b) Verify that BCT = 0 and deduce the fundamental cutset matrix from the matrix A. (c) Compute Ybus matrix with ground as the reference. (d) Choosing 6 as the reference bus, compute Zbus using the procedure of Section 3.7.4. (e) In the Ybus derived in step 3, since the ground bus is isolated, delete the row and column corresponding to the ground bus to get a reduced Ybus. Verify that this reduced Ybus is the inverse of Zbus obtained in step 4.

Topological Analysis of Power Networks

3.41

3.7 In the Ybus of Problem 3.6, eliminate the nodes 3 and 5 and obtain the reduced admittance matrix with ground bus as the reference. 0

0 1

5

2

4

3

3.8 The system impedance data for the system in Fig. P.3.5 is given in the Table P.3.1. Lines 1-0 and 2-0 represent the transient reactances of the generators connected at buses 1 and 2, respectively.

From Bus

To Bus

R

X

1

4

0.15

0.60

1

5

0.05

0.20

2

3

0.05

0.20

2

4

0.10

0.40

2

5

0.05

0.20

3

4

0.10

0.40

1

0

—

0.25

2

0

—

1.25

the given Ybus. (b) Implement the program for the above system to eliminate all the buses except 7 and 8, and obtain a reduced bus admittance matrix at buses 7 and 8. 3.9 For the system in Problem 3.8, neglect all series resistance and also omit the line 2-4. (a) Compute Zbus with ground as the reference bus. (b) Compute Ybus with ground as the reference. (c) Verify that ZbusYbus = U.

4 As discussed in Chapter 3, the bus impedance matrix of a transmission network is a mathematical equivalent through which the bus voltages are explicitly expressed in terms of the bus currents, giving rise to a multiport representation of the transmission network. The primitive impedance matrix and the network topology are required to construct the bus impedance matrix Zbus. The elements of the bus impedance matrix are the open-circuit impedance parameters of the multiport representation. The topological formula for the computation of Zbus, however, requires the inversion of a matrix whose order is equal to the number of independent loops in the network and hence, the method cannot be used to compute Zbus for large networks. An alternate method of constructing the bus impedance matrix by adding one element at a time has been very popular with power engineers, and is known as the Zbus building algorithm. In this chapter, we shall develop the algorithm from topological considerations and illustrate it by means of examples.

A partial network is a connected network consisting of some or all of the buses, and some of the lines of a given transmission network. Let z0, A0, and Zbus(a) be the primitive impedance matrix, bus incidence matrix and the bus impedance matrix, respectively, of the partial network shown in Fig. 4.1. When certain

1 2 Partial Network

line(s) or removal of certain line(s) in the network are made, we would like matrix Zbus of the new network in terms of Zbus(0).

m

0

Partial network (interconnections are not shown)

4.2

Computer Techniques in Power System Analysis

For the partial network, using the expression for the bus admittance matrix (3.73), we have Z bus(0) = ÈÎYbus(0) ˘˚

-1

= ÈÎ A 0 z 0-1 AT0 ˘˚

-1

(4.1)

Suppose an element a having self-impedance Zaa and mutual impedance z0a with the elements of the partial network, is added to the partial network. Then for the new network, the element voltage and current vectors respectively are Èi0 ˘ Èv0 ˘ v = Í ˙ and i = Í ˙ Îva ˚ Îia ˚ where v0 and i0 are the element voltage and current vectors respectively in the partial network, and va and ia are the voltage and current variables respectively of the added element. The primitive equations in impedance form are È v0 ˘ È z0 Ív ˙ = Í z Î a ˚ Î a0

z0a ˘ È i0 ˘ zaa ˙˚ ÍÎia ˙˚

(4.2)

and in the admittance form they are given by È i0 ˘ È y00 Íi ˙ = Í y Î a ˚ Î a0 where

y0a ˘ È v0 ˘ yaa ˙˚ ÍÎva ˙˚

-1 z0-1 = y00 - y0a yaa ya 0

(4.3) (4.4)

Note that y00 π z0-1 unless y0a = ya 0 , i.e., the added element is not coupled to any of the elements of the partial network. Combining (4.2), (4.3), and (4.4), the primitive equations can be written in a mixed form as -1 Èi0 ˘ È z0 = Í Ív ˙ -1 ya 0 Î a ˚ ÍÎ- yaa

-1 ˘ Èv0 ˘ y0a yaa ˙Í ˙ -1 yaa ˙˚ Îia ˚

(4.5)

If Vbus and Ibus denote the bus voltage and the bus current vectors corresponding to the multiport description of the new network, then from Eqs. (3.59) and (3.71) (Chapter 3), we get È v0 ˘ v = Í ˙ = AT Vbus Îva ˚

(4.6)

È i0 ˘ I bus = Ai = A Í ˙ (4.7) Îia ˚ where A is the bus incidence matrix of the new network. With these preliminaries, we

Bus Impedance Algorithm

Let the element a be added between the buses p and q, with the orientation from p to q as shown in Fig. 4.2. Since the addition of an element connecting two nodes creates a closed path in the partial network, we designate such an element as a link. This is slightly different from the

1 2 p Partial Network

it constituted an element of the co-tree. We retain the term link to be consistent with the prevailing terminology. When the link is added, no new bus is created. The bus incidence matrix A for the new network can be partitioned as A = [ A0

K1 ]

4.3

a q m 0

(4.8)

New network showing addition of a link

where, K1 = [0 � 1 0 � 0 -1 0 � 0]T K1 is the sub-matrix of the bus incidence matrix corresponding to the added element. It essentially indicates how the new element is incident to the partial network. Hence, K1 has only two non-zero entries, +1 and -1 in the pth row and the qth row, respectively. Making use of Eq. (4.8) in Eqs. (4.6) and (4.7), the bus variables and primitive variables for the new network are related as v0 = A0T Vbus

(4.9)

va = K1T Vbus

(4.10)

A0 i0 + K1 ia = I bus

(4.11)

Now eliminating the variables v0, i0, va, and ia from the set of equations (4.5), (4.9), (4.10), and (4.11), an explicit relation between Vbus and Ibus is obtained as -1 ÈÎ B11 - B12 B22 B21 ˘˚ Vbus = I bus

where

È B11 ÍB Î 21

A0 z0-1 A0T B12 ˘ È = Í -1 B22 ˙˚ ÍÎ- K1T + yaa ya 0 A0T

(

)

-1 ˘ K1 + A0 y0a yaa ˙ -1 yaa ˙˚

(4.12)

(4.13)

Hence, the bus impedance matrix (Zbus) of the new network can be written as -1 Z bus = ÈÎ B11 - B12 B22 B21 ˘˚

-1

(4.14)

4.4

Computer Techniques in Power System Analysis

To evaluate the inverse on the right hand side of Eq. (4.14), the following special matrix inversion formula is used. -1

[ F + GHK ]-1 = - F -1G ÎÈ H -1 + KF -1G ˘˚ KF -1 + F -1

(4.15)

Using Eq. (4.15) in Eq. (4.14), we get -1 -1 -1 -1 Z bus = B11 + B11 B12 ÎÈ- B21 B11 B12 + B22 ˘˚ -1 B21 B11

(4.16)

Since B11 = A0 z0-1 A0T , using Eq. (4.1), -1 B11 = ÈÎ A0 z0-1 A0T ˘˚

-1

= Z bus(0)

(4.17)

Hence, the expression for Zbus in Eq. (4.16) reduces to Zbus = Zbus(0) – Zbus(0)C12D–1C21 Zbus(0)

(4.18)

where C12 = K1 +

A0 y0a yaa

(4.19)

C 21 = K1T +

ya 0 A0T yaa

(4.20)

D = C 21 Z bus (0)C12 +

1 yaa

(4.21)

When there is no mutual coupling between the added element and the elements in the partial network, both y0a and ya0 become null vectors and hence, the bus impedance matrix of the new network is given by Z bus = Z bus(0) -

Z bus(0) K1 K1T Z bus(0) K1T Z bus(0) K1 + zaa

(4.22)

where Zaa is the primitive impedance of the added element. It may be noted that C12 and C21 in Eqs. (4.19) and (4.20) need not be the transpose of each other. If a phase shifting component is the added element, then y0a π y aT 0 and hence C12 π C T21. Phase shifting components such as the tap changing transformer with a complex turns ratio have a primitive admittance matrix of order 2 ¥ 2 which is asymmetric as well as singular. The presence of such asymmetry in the primitive admittance matrix will make the bus impedance matrix asymmetric. So far, no assumption has been made regarding the symmetry of the primitive impedance and the bus impedance matrices. Hence, Eqs. (4.18) and (4.22) are applicable even when the added element is a phase-shifting component. However, the singular nature of the primitive admittance matrix calls for a special computational procedure. For further discussion, unless otherwise stated, it is assumed that the added element is not a phase-shifting component.

Bus Impedance Algorithm

4.5

Since the added element is not a phase-shifting component, therefore, T y0a = yaT 0 and C12 = C 21

Z bus = Z bus(0) where

Z bus(0)C1C1T Z bus(0) C1T Z bus(0)C1 + (1 yaa )

C1 = K1 +

A0 y0a yaa

(4.23)

(4.24) (4.25)

When there is no mutual coupling between the added link and the elements in the partial network, Eq. (4.23) reduces to Eq. (4.22). As a special case, if the added link is between the reference bus and the bus q, then the vector K1 reduces to K1 = [0 � 0 -1 0 � 0]T This vector K1 has to be used in the appropriate equations.

(4.26)

When a number of links are added at a time, following the procedure similar to that of the addition of a single link, we get -1

where

Z bus = Z bus(0) - Z bus(0)C1 z� C1T Z bus(0)

(4.27)

-1 z� = C1T Z bus(0)C1 + yaa

(4.28)

-1 C1 = K1 + A0 y0a yaa (4.29) K1 is the bus incidence matrix corresponding to the added links. y0a and yaa are also matrices here.

When the added links are not mutually coupled with any element in the partial network, Eq. (4.27) reduces to -1

where

Z bus = Z bus(0) - Z bus(0) K1 z� K1T Z bus(0)

(4.30)

z� = K1T Z bus(0) K1 + Zaa

(4.31)

in which zaa is the primitive impedance matrix of the added elements.

Let the element α be added at the bus p creating a new bus q with the orientation from p to q, as shown in Fig. 4.3. This is termed as the addition of a branch. The word

4.6

Computer Techniques in Power System Analysis

“branch” used here has a slightly different connotation from that used in Chapter 3 terminology because of its widespread use in the literature while discussing bus impedance algorithm. 1 2 a

p Partial Network

q

m

0

New network owing to the addition of a branch

The bus incidence matrix for the new network is given by È Í A0 A=Í Í Í Î0 � where,

˘ ˙ K2 ˙ ˙ ˙ 0 -1 ˚

(4.32)

K 2 = [0 � 0 1 0 � 0]T

(4.33)

For the new network, let the vectors Vbus and Ibus in the multiport description be partitioned as follows: Let Vbus ¢ , and I bus ¢ be the variables corresponding to all the buses except the newly added bus q, whose variables are denoted by Vq and Iq. Thus T

T

Vbus = ÈÎVbus I q ˘˚ ; ¢ Vq ˘˚ ; I bus = ÈÎ I bus ¢ Substituting (4.32) and Eq. (4.34) in Eqs. (4.6) and (4.7), we get

(4.34)

v0 = A0T Vbus ¢

(4.35)

va = K 2T V ¢ bus - Vq

(4.36)

A0 i0 + K 2 ia = I ¢ bus

(4.37)

ia = - I q

(4.38)

The primitive equations for the new network are given by Eq. (4.5). To obtain the relation between Vbus and Ibus, the other variables v0, i0, va, and ia are to be eliminated from Eq. (4.5), and Eqs. (4.35) – (4.38).

Bus Impedance Algorithm

4.7

Elimination of these variables yields (details omitted) È A0 z0-1 A0T Í -1 ÍÎ- K 2T + yaa ya 0 A0T

(

)

(

)

-1 ˘ ÈV ¢ bus ˘ È I bus - K 2 + A0 y0a yaa ¢ ˘ ˙Í =Í ˙ ˙ -1 ˙˚ Î I q ˚ Î-Vq ˚ - yaa

(4.39)

-1 , we get Rearranging Eqs. (4.39) and using the relation A0 z0-1 A0T = Z bus(0)

ÈV ¢ bus ˘ È Z bus(0) Í V ˙ = ÍD Z Î q ˚ Î 21 bus(0) where

¢ ˘ ˘ ÈI bus ˙Í I ˙ D21 Z bus(0) D12 + (1 yaaa )˚ Î q ˚ Z bus(0) D12

D12 = K 2 +

A0 y0a yaa

(4.40)

(4.41)

ya 0 A0T yaa Hence, the bus impedance of the new network is given by D21 = K 2T +

(4.42)

Z bus (0) D12 ˘ È Z bus (0) Z bus = Í (4.43) ˙ Î D21Z bus (0) D21Z bus ( 0) D12 + (1 yaa )˚ It should be noted that D12 and D21 in Eqs. (4.41) and (4.42) need not be the transpose of each other. Therefore, Eqs. (4.40) and (4.43) are very general and can be used even if the added element is a phase-shifting component. However, to simplify matters for subsequent discussions, it is assumed that the added element is not a phase-shifting component, so that y0a = yaT 0 .

T D12 = D21 = C2 (say ) . Hence, Eq. (4.43) reduces to

È Z bus (0) Z bus = Í T ÍÎC 2 Z bus (0) where

˘ ˙ + (1 yaa )˙˚

Z bus (0) C 2 C 2T Z bus (0) C 2

C2 = K 2 +

A0 y0a yaa

(4.44) (4.45)

Addition of a Branch with no Mutual Coupling When there is no mutual coupling between the added element and the elements in the partial network, Eq. (4.44) Z bus(0) K 2 È Z bus(0) ˘ Z bus = Í T (4.46) ˙ T ÍÎ K 2 Zbus(0) K 2 Z bus(0) K 2 + zaa ˙˚ where zaa is the primitive impedance of the added element. As a special case, the added branch may be from the reference bus. In such a case, the vector K2 reduces to a null vector. The null vector has to be used in the appropriate equations. A similar procedure can be followed when a number of branches are added at a time. In such a case,

4.8

Computer Techniques in Power System Analysis

È Z bus(0) Z bus = Í T ÍÎC 2 Z bus(0)

Z bus(0) C 2 C 2T Z bus(0) C 2

+

-1 yaa

˘ ˙ ˙˚

(4.47)

where the matrix -1 C 2 = K 2 + A0 y0a yaa

(4.48)

in which K2 is the sub-matrix of the bus incidence matrix corresponding to the added branches. When the added branches are not mutually coupled with any element in the partial network, Eq. (4.47) reduces to È Z bus(0) Z bus = Í T ÍÎ K 2 Z bus(0)

˘ ˙ + zaa ˙˚

Z bus(0) K 2 K 2T Z bus(0) K 2

(4.49)

in which zaa is the primitive impedance matrix of the added elements.

Removal of a link from a network means the removal of the self and mutual admittances corresponding to the element from the primitive admittance matrix of

have the self and mutual admittances of values exactly equal to those of the link to be removed, but with a negative sign; and its mutual admittance with the link to be removed should be zero. Now the same algorithm for the addition of a link

to be removed, it may be noticed that the product of the bus incidence matrix of the equal to –A0y0α , where A0 is the bus incidence matrix of the network with element α removed, and y0α is the vector showing the mutual admittance between the element α element is -yαα. Therefore, the vector C1 can still be computed from Eq. (4.25). Thus, when a link α is removed from the network whose bus impedance matrix is Zbus(0), Zbus is given by Z bus = Z bus(0) -

where

Z bus(0)C1C1T Z bus(0) C1T Z bus (0)C1 - (1 yaa )

C1 = K1 +

A0 y0a yaa

(4.50)

(4.51)

If the link to be removed is not mutually coupled with any other element in the network, Eq. (4.50) reduce to

Bus Impedance Algorithm

Z bus = Z bus(0) -

Z bus(0) K1 K1T Z bus(0) K1T Z bus(0) K1 - zaa

4.9

(4.52)

where zaa is the primitive impedance of the element to be removed. Extending the method of removal of a single link (Eq. (4.50)), the bus impedance matrix for the removal of more than one link is obtained as

where

-1 Z bus = Z bus(0) - Z bus(0)C1 z� C1T Z bus(0)

(4.53)

-1 z� = C1T Z bus(0)C1 - yaa

(4.54)

-1 C1 = K1 + A0 y0a yaa

(4.55)

It may be noted that K1, yαα , and y0α in Eqs. (4.53) – (4.55) are no longer vectors, but matrices. When the links to be removed are not mutually coupled with any element in the partial network, Eq. (4.53) reduces to -1

where

Z bus = Z bus(0) - Z bus(0) K1 z� K1T Z bus(0)

(4.56)

z� = K1T Z bus(0) K1 - zaa

(4.57)

in which zaa is the primitive impedance matrix of the links to be removed.

Consider the case of the addition of a link. If the added element has no mutual coupling, the vector y0a becomes a null vector. In such a case, while computing A0 y0a in Eq. (4.24), the bus incidence matrix A0 need not be considered at all. If the added element is mutually coupled, it is enough to consider the elements in the coupled group alone, since the components of y0a corresponding to the elements that are not in the couple group, are zero. While computing A0 y0a, the reduced vector y0a denoted by y0a, and the reduced matrix A0 denoted by A0, can be used. The rows of y0a and the columns of A0 correspond to the elements having mutual coupling with the added element α y0a and yaa is now limited to the number of elements in the coupled group. All these remarks are applicable not only in the case of the addition of a link but also for all other cases discussed. A word of caution is necessary regarding the orientations of the elements and the sign of the mutual impedance in the primitive impedance matrix of the coupled group. In circuit theory, the proper dot convention is normally followed to specify the sign of mutual impedances. But in power systems, mutual coupling arises when two transmission lines run parallel for a considerable distance. The sign of mutual impedance depends on the orientations assigned to them in the oriented graph. If the

4.10

Computer Techniques in Power System Analysis

elements have like orientations when emanating from a bus, the mutual impedance will have a positive value; and if the element shave opposite orientations, the corresponding mutual impedance will have a negative value. All these points are brought out in the solved numerical examples at the end of the chapter.

When an element corresponding to a radial line is removed, one bus gets isolated and the number of buses in the network is reduced by one. From Eq. (4.44) it can be can be obtained from the original bus impedance matrix by simply deleting the row and the column corresponding to the isolated bus. If the isolated bus is the reference

When the parameter of an element is changed, the bus impedance matrix can be an element with the revised parameter. This particular case has been illustrated through a numerical example (Example 4.3).

In this chapter, a generalized bus impedance algorithm has been developed based on certain fundamental concepts of the graph theory. The bus incidence matrix which shows how the various elements are interconnected, is essentially used at various stages. In other words, this development exploits the topology of the network. The merits of this algorithm are summarized below. 1. The algorithm for the addition of a link is a one-step procedure instead of a two-step procedure as in other algorithms. 2. Mutual couplings between the elements of the network do not pose any problem either at the addition or the removal stage. 3. The algorithm is amenable for the addition of the link and the branch either one at a time or in groups. 4. The algorithm is comprehensive so as to include the effects of phase-shifting components, changes in network parameters, etc. Compute the bus impedance matrix for the network shown in Fig. 4.4. Bus 1 can be taken as the reference bus, since there are no shunt connections to the ground in this case. The oriented linear graph for the network is shown in Fig 4.5 Step 1: Element a is added. It is a branch from the reference bus. Then,

Bus Impedance Algorithm

4.11

a Z bus = 2 [0.1] 5

2

4

e(0.2)

f (0.3)

0.1

0.2 b (0.4)

3 c (0.5)

a (0.1)

d(0.5)

1

f

e

5

4

2 b a

3 d

c 1

Step 2: Element b is added. It is a branch from bus 2 to bus 3 and has no mutual coupling with the element in the partial network. b K 2 = 2 [1]; zaa = 0.4 Using Eq. (4.46), we get 2 3 2 È0.1 0.1˘ Z bus = Í 3 Î0.1 0.5˙˚ Step 3: Element c is added. It is a link between the reference buses 1 and 3, and has no mutual coupling with elements in the partial network.

4.12

Computer Techniques in Power System Analysis

c K1 =

Thus,

2 È0˘ ; z = 0.5 3 ÍÎ-1˙˚ aa

È0.1 0.1˘ È 0 ˘ È - 0.1˘ Z bus(0) K1 = Í ˙Í ˙=Í ˙ Î0.1 0.5˚ Î-1˚ Î- 0.5˚ È - 0.1˘ K1T Z bus(0) K1 = [0 -1] Í ˙ = 0.5 Î- 0.5˚ From Eq. (4.22) È0.1 Z bus = Í Î0.1 È0.1 =Í Î0.1

0.1˘ Ê È - 0.1˘ ˆ [- 0.1 - 0.5]˜ (0.5 + 0.5) ¯ 0.5˙˚ ÁË ÍÎ- 0.5˙˚ 0.1˘ È 0.01 0.05 ˘ 0.5˙˚ ÍÎ0.05 0.025˙˚

2 3 2 È0.09 0.05˘ = Í 3 Î0.05 0.25˙˚ Step 4: Element d is added. It is a branch from the reference bus and has no mutual coupling with the elements in the partial network. d 2 È0 ˘ K 2 = Í ˙ ; zaa = 0.5 3 Î0 ˚ Using Eq. (4.46), we get 2 3 4 2 È0.09 0.05 0.0˘ Z bus = 3 Í0.05 0.25 0.0˙ Í ˙ 4 ÎÍ 0.0 0.0 0.5˙˚ Step 5: Element e is added. It is a branch from bus 2 creating a new bus 5. It has mutual coupling with the existing element b. e 2 È1 ˘ K 2 = 3 Í0 ˙ Í ˙ 4 ÍÎ0˙˚ Let zc and yc be the primitive impedance and the admittance matrices of the coupled group. Then, b e b È0.4 0.1˘ zc = Í e Î 0.1 0.2˙˚

Bus Impedance Algorithm

4.13

Inverting zc we get, b e b È 2.8570 -1.4285˘ yc = Í e Î-1.4285 5.7140 ˙˚ This gives a c d

b

y0a = [0 0 0 -1.4285]T ;

yaa = 5.7140;

A0 is given by a c d b 1˘ 2 È-1 0 0 A0 = 3 Í 0 -1 0 -1˙ Í ˙ 4 ÎÍ 0 0 -1 0˙˚ Clearly we can see that for calculating the product A0 y0α , the inversion of the complete primitive impedance matrix and the use of the complete A0 matrix is not necessary. Only the column of A0 corresponding to the branch b (denoted by A–0) and the element of y0α corresponding to the branch b (denoted by y0a – Using A0 and y0a in Eq. (4.45), we get È1 ˘ È 1 ˘ È0.75˘ 1 Í ˙ Í ˙ Í C 2 = 0 + -1 [ -1.4285] = 0.25˙ Í ˙ Í ˙ ˙ 5.7140 Í ÍÎ0˙˚ ÍÎ 0˙˚ ÍÎ0.000˙˚ È0.09 0.05 0.00˘ È0.775˘ È0.08˘ Z bus(0)C 2 = Í0.05 0.25 0.00˙ Í0.25˙ = Í0.10˙ Í ˙Í ˙ Í ˙ ÍÎ0.00 0.00 0.50˙˚ ÍÎ0.00˙˚ ÍÎ0.00˙˚ C 2T Z bus(0)C 2

È0.08˘ = [0.75 0.25 0.00] Í0.10˙ = 0.085 Í ˙ ÍÎ0.00˙˚

C 2T Z bus(0)C 2 + (1 yaa ) = 0.085 + (1 5.7140) = 0.26 Hence, using Eq. (4.44),

Z bus

a È0.09 Í0.05 = Í Í0.00 Í Î0.08

c 0.05 0.25 0.00 0.10

d 0.00 0.00 0.50 0.00

b 0.08˘ 0.10˙ ˙ 0.00˙ ˙ 0.26˚

Step 6: Element f is added. It is a link between buses 4 and 5 and has mutual coupling

4.14

Computer Techniques in Power System Analysis

with element b. f 2 È 0˘ 3 Í 0˙ ; K1 = Í ˙ 4 Í 1˙ Í ˙ 5 Î -1˚

b zc =

f

b

b È0.4 0.2˘ ; f ÍÎ0.2 0.3˙˚

yc =

f

b È 3.75 -2.5˘ f ÍÎ-2.5 5.00 ˙˚

b 2 È 1˘ 3 Í -1˙ ; A0 = Í ˙ 4 Í 0˙ Í ˙ 5 Î 0˚

y0a

b = [ -2.5] ;

yaa = 5.00 ;

Hence, using Eq. (4.25)

C1 = K1 +

A0 y0a yaa

È-0.50˘ Í 0.50 ˙ ˙ =Í Í 1.00 ˙ Í ˙ Î-1.00˚

Using Eq. (4.24), we get,

Z bus

È0.080 Í0.050 = Z bus(0) - T =Í C1 Z bus(0)C1 + (1 yaa ) Í0.050 Í Î0.0555

0.055 ˘ 0.100 ˙ ˙ 0.125 ˙ ˙ 0.100 0.125 0.1975˚

0.050 0.050 0.250 0.000 0.000 0.250

Z bus(0)C1C1T Z bus(0)

bus impedance matrix when the self impedance of the line f is changed from 0.3 to 0.24, and also when the mutual impedance between the line b and f is changed from 0.2 to 0.0. g e 2

5

4 f¢

b

3

This case can be treated as the simultaneous removal of the line f with f ¢) and the addition of a line g with the new impedance value. Thus this case belongs f ¢ has the

Bus Impedance Algorithm

4.15

self and the mutual admittances as the negative of the link to be removed, and has no mutual admittance with the link to be removed. Link g is not mutually coupled to the other elements. Fig. 4.6 shows the oriented graph of the coupled group. The primitive admittance matrix of the coupled group is obtained as b

e

f¢

g

b È 4.6153 -2.3077 -3.0769 0.0 ˘ ˙ e Í-2.3077 6.1538 1.5385 0.0 Í ˙ f ¢ Í-3.0769 1.5385 -5.3846 0.0 ˙ Í ˙ g Î 0.0 0.0 0.0 4.1667 ˚ È-3.0769 0.0˘ È-5.3846 0.0 ˘ y0a = Í ; yaa = Í ˙ 4.1667 ˙˚ Î 1.5385 0.0˚ Î 0.0 Also, from the oriented graph in Fig. 4.6, b e f¢ g 2 È 1 1˘ 2 È 0 0˘ 3 Í-1 0˙ 3 Í 0 0˙ ˙ ; K1 = Í ˙ A0 = Í 4 Í 0 0˙ 4 Í 1 1˙ Í ˙ Í ˙ 5 Î 0 -1˚ 5 Î-1 -1˚ For this case,

-1 yaa

f¢ g f ¢ È-0.186 0.0 ˘ = Í g Î 0.0 0.240˙˚

-1 Substituting the values of y0a , yaa , A 0 , , and K1 in Eq. (4.29), C1 is obtained as

È 0.2855 0.0˘ Í-0.5715 0.0˙ ˙ C1 = Í 1.0˙ Í 1.0 Í ˙ Î-0.7140 -1.0˚ -1 � Knowing Zbus(0) in Example 4.1, C1 and yaa , z is obtained from Eq. (4.28) as

È- 0.0369 0.1428˘ ˙ Î 0.1428 0.4485˚

z� = Í Hence,

z�

-1

È12.230 3.890 ˘ ˙ Î 3.890 -1.007 ˚

=Í

4.16

Computer Techniques in Power System Analysis

-1 Finally, substituting the values of Zbus(0) and the calculated C1 and z� in Eq. (4.27),

Z bus

2 2 È0.0836 3 Í0.0420 = Í 4 Í0.0400 5 ÍÎ0.0592

3

4

5

0.0420 0.0400 0.0592˘ 0.2400 0.5000 0.0740˙ ˙ 0.5000 0.2500 0.1300˙ ˙ 0.0740 0.1300 0.1920˚

4.1 Using the building algorithm, construct the bus impedance matrix Zbus for the network in Fig. P.4.1. Choose bus 3 as the reference bus. The system data is given below: From bus 1 1 2 2 2 3

To bus 2 3 3 4(1) 4(2) 4

X 0.5 0.5 0.6 0.4 0.4 0.5

Mutual coupling between 2-4(1) and 2-4(2) may be taken as 0.1 1

2

3

2 0 .2 1.0

(1)

(2)

0 .5

0.3

0 .3 3

4

4

1

4.2 Using the building algorithm, construct Zbus for the system in Fig. P.4.2. Choose 4 as the reference bus. 4.3 Modify the Zbus of Problem 4.1 when (a) line 1-2 is removed, (b) line 2-4(1) is removed, (c) mutual impedance between lines 2-4(1) and 2-4(2) is changed to 0.15, and (d) another line is added between buses 1 and 2 having self-impedance of j0.4. 4.4 Modify Zbus of Problem 4.2 if the mutual reactance between lines 1-3 and 2-3 is absent.

5 When an abnormal condition arises in a power system such as a fault, an insulator

circuit studies, a number of assumptions are made which simplify the formulation of

5.2

Computer Techniques in Power System Analysis

È zs Íz Í m ÍÎ zm

zm zs zm

zm ˘ zm ˙ ˙ zs ˚˙

Short Circuit Studies

ÈV a ˘ È z Í b˙ Í s ÍV ˙ = Í zm Í c ˙ Íz ÎÍV ˚˙ Î m

a zm ˘ È I ˘ Í ˙ zm ˙ Í I b ˙ ˙ zs ˚˙ ÍÍ I c ˙˙ Î ˚

zm zs zm

V a , b, c = z a , b, c V S = S3V a ,b,c ;

I

a , b, c

I S = S3 I a , b , c ;

where È1 1 1 Í S3 = Í1 a 3Í 2 Î1 a

1˘ ˙ a2 ˙ ; a ˙˚

a = 1–120∞ = -0.5 + j 0.866

S property that S3-1

=

( )

T S3*

È1 1 1 Í 2 = Í1 a 3Í Î1 a

1˘ ˙ a˙ a 2 ˙˚

S S3V a ,b,c = S3 z a ,b,c S3-1 S3 I a ,b,c V s = zs I s z S = S3 z a ,b,c S3-1 ;

where zs turns out to be È z s + 2 zm S z =Í 0 Í ÍÎ 0

0 z s - zm 0

˘ Èz ˙=Í0 ˙ Í zs - zm ˙˚ ÍÍ 0 Î 0 0

0

0 z1 0

0˘ ˙ 0˙ ˙ z 2 ˙˚

5.3

5.4

Computer Techniques in Power System Analysis

z0, z1, and z2 zs by the symbol z0,1,2 za,b,c V s and I s are denoted by V 0,1,2 and I

0,1,2

È Va +Vb +Vc ˘ ÈV 0 ˘ Í 1˙ 1 Í a ˙ 2 c b = ÍV ˙ = ÍV + aV + a V ˙ 3Í a Í 2˙ c˙ 2 b ÍÎV ˙˚ ÍÎV + a V + aV ˙˚

V 0,1, 2

I 0,1, 2

È Ia + Ib + Ic ˘ ÈI 0 ˘ Í 1˙ 1 Í a ˙ 2 c b = ÍI ˙ = Í I + aI + a I ˙ 3Í a Í 2˙ c˙ 2 b ÍÎ I ˙˚ ÍÎ I + a I + aI ˙˚

V 0 = z 0 I 0 ; V 1 = z1 I 1 ; V 2 = z 2 I 2 ;

I0

z0

I1

V0 z1

I2

V1 z2

+ + +

z

-

V2

a ,b , c

È zsaa = Í zmab Í ÍÎ zmac

zmab zsbb zmbc

zmac ˘ zmbc ˙ ˙ zscc ˙˚

Short Circuit Studies

5.5

z S = S3 z a ,b,c S3-1

È (z + z + z ) ( zsaa + a 2 zsbb + azscc ) ( zsaa + azsbb + a 2 zscc ) ˘ saa Sbb Scc Í ˙ - ( zmbc + azmab + a 2 zmac ) - ( zmbc + azmac + a 2 zmab ) ˙ Í + 2 ( zmab + zmbc + zmac ) Í ˙ Í ˙ Í z + az + a 2 z ˙ 2 + + z a z az ) ) (Í saa sbb ( scc saa sbb scc ( zsaa + zsbb + zscc ) ˙ 1 = Í- ( zmbc + azmac + a 2 zmab ) - ( zmab + zmbc + zmac ) + 2 ( zmbc + azmab + a 2 zmac )˙ ˙ 3Í ˙ Í Í ˙ Í( zsaa + a 2 zsbb + azscc ) ˙ ( zsaa + azsbb + a 2 zscc ) ( zsaa + zsbb + zscc ) Í ˙ 2 2 - ( zmab + zmbc + zmac ) ˙ Í- ( zmbc + azmab + a zmac ) + 2 ( zmbc + azmac + a zmab ) Í ˙ ÍÎ ˙˚

zsaa = zsbb = zscc and zmab = zmac = zmbc ya,b,c, then

y 0,1,2 = S3 y a ,b,c S3-1 È y0 Í =Í0 Í ÍÎ 0 where

y0 =

1 z1

, y1 =

1 z1

, and y 2 =

1 z2

0 1

y

0

0˘ ˙ 0˙ ˙ y 2 ˙˚

.

V a as

V a = V a –0 ; V b = V a – - 120 ; V c = V a – - 240 θ

V a and I a I a = I a –q ; I b = I a – (q - 120) ; I c = I a – (q - 240)

5.6

Computer Techniques in Power System Analysis

V 0 = 0; V1 = 3V a ; V 2 = 0 I 0 = 0; I1 = 3 I a ; I 2 = 0

V 1 = z1 I 1

V 1 = 3V a ; I 1 = 3I a ; I a =

Va z1 a

a

zF g

Ia + Ib + Ic a zF

zg

c

b zF

zF n zg 0 (a)

a + b Va +

Ia Ib

zF

Ia

a + b V +

Ib

zF

a

Ic c b V + Vc 0 - - -

V c + Vc - - - 0

Ic

zF

b

(b)

zg

(c)

Short Circuit Studies

ÈV a ˘ È z F + z g Í b˙ Í ÍV ˙ = Í z g Í c˙ Í ÍÎV ˙˚ Î z g

zg

ÈI a ˘ Í b˙ ÍI ˙ Í c˙ ÍÎ I ˙˚

˘ ˙ zg ˙ ˙ zF + z g ˚ zg

zF + z g zg

5.7

V a ,b , c = Z a ,b , c I a ,b , c Z a,b,c yields

Z

0,1, 2

Z

Y 0,1, 2

È z0 Í =Í0 Í ÍÎ 0

0 ˘ È zF + 3z g ˙ Í 0˙=Í 0 2˙ Í 0 z ˙˚ Î

0 1

z

0

zF 0

0˘ ˙ 0˙ z F ˙˚

zF and is independent of zg Y 0,1,2

0,1,2

È y0 Í =Í0 Í ÍÎ 0

0

0 y1 0

0 ˘ ÈÍ( z F + 3 z g ) ˙ 0˙=Í 0 Í 2˙ y ˙˚ Í 0 Î y1 =

-1

0

( zF-1 ) 0

0 ˘ ˙ 0 ˙ ˙ z F-1 ˙ ˚

( )

1 = yF zF

zg Æ • I a ,b , c

= Y a,b,cV a,b,c

where

Y

a ,b , c

= (Z

)

a ,b ,c -1

È yt + 2 yF 1Í = Í yt - yF 3 ÍÎ yt - yF

with

yt =

1 zF + 3z g

yt - yF yt + 2 yF yt - yF

yt - yF ˘ yt - yF ˙ ˙ yt + 2 yF ˙˚

5.8

Computer Techniques in Power System Analysis

a

c

b

a + zF

zF

b Va +

zF

Ia Ib

Ic c Vb + Vc 0 - - -

0 (a)

(b)

YFa ,b,c zg = •, then Y a,b,c becomes Y a ,b , c

Y

0,1, 2

È0 = ÍÍ0 ÍÎ0

È 2 yF

- yF

- yF ˘

ÍÎ- yF

- yF

2 yF ˙˚

1 = ÍÍ- yF 2 yF - yF ˙˙ 3

0 yF 0

0 0 ˘ Èy Í 0 ˙=Í0 ˙ yF ˙˚ ÍÍ 0 Î

0 1

y

0

0˘ ˙ 0˙ ˙ y 2 ˙˚ Zg = •

yF

zF yF

n Zbus m buses are the buses where

Short Circuit Studies

5.9

m ¢

¢ I1

1 0¢

1

+ V1 +

2

I2

2

V2

m

In + Va 0 -

n

n

+ Vn - - -

0 (a)

0 (b)

V0a

¢ Zbus can be obtained with the

Vbus = Zbus Ibus ¢ V0a

n

V0a Vbus = Z bus I bus + bV0a

where b = [1 1

1]T

pth VE = zF IF th

p

5.10

Computer Techniques in Power System Analysis

are written from inspection as I p ( F ) = - I F ; V p ( F ) = VF ; I i ( F ) = 0 ; Vi ( F ) = unknown for i = 1, 2, ... n and i π p

where Ip F and Vp F

1 2

p +

Ip(F)

IF + zF

VF

p

IF + Ip(F)

n

VF -

-

-

0

0 (a)

(b)

n V1( F ) = Z11 I1( F ) +

+ Z1 p I p ( F ) +

+ Z1n I n ( F ) + V0a

V2( F ) = Z 21 I1( F ) +

+ Z2 p I p( F ) +

+ Z 2 n I n ( F ) + V0a

V p (F ) = Z p1 I1( F ) +

+ Z pp I p ( F ) +

+ Z pn I n ( F ) + V0a

Vn (F ) = Z n1 I1( F ) +

+ Z np I p ( F ) +

+ Z nn I n ( F ) + V0a pth

z F I F = - Z pp I F + V0a

fi IF =

V0a Z pp + z F

VF = V p ( F ) = z F I F = z F ( Z pp + z F ) V0a . -1

(

ViF = V0a - Z ip Z pp + z F

)

-1

V0a i = 1, 2, … , n iπ p

Short Circuit Studies

5.11

description is IF = yFVF VF,

we now substitute for IF in the pth V p (F ) = - Z pp yF V p ( F ) + V0a

(

V p (F ) = 1 + Z pp yF

)

-1

V0a

I F = yF (1 + Z pp yF ) V0a -1

and

pth Vi ( F ) = V0a + Z ip I p ( F ) = V0a - Z ip I F = V0a - Z ip yF (1 + Z pp yF ) V0a -1

IF = yF VF

formulation for the admittance description of a fault has been included because

Z

Y

Zbus Zbus Ybus For Ybus V0a z j in parallel with an impedance z j = ( j = 1, 2, … , m) of the n ÈV a I bus = YbusVbus - I 0a where, I 0a = Í 0 Î z1

V0a z2

V0a zm

0 0

˘ 0˙ ˚

T

5.12

Computer Techniques in Power System Analysis

ViF i = 1, 2, …, n i π p ViF i = 1, 2, …, n n

Ip F

n Zbus Zbus Ybus can be Zbus Zbus and Ybus

ÈV1 ˘ ÍV ˙ = Í 2˙ ÎÍV3 ˚˙

j

È0.1274 0.1061 0.0981˘ È I1 ˘ È1˘ Í 0.1061 0.1345 0.1151˙ Í I ˙ + Í1˙ V a Í ˙ Í 2˙ Í ˙ 0 ÍÎ 0.0981 0.1151 0.1215˙˚ ÍÎ I 3 ˙˚ ÍÎ1˙˚

j0.06 j0.06

j0.20

2 3 j0.08

0¢

I1 V0

+ V1

j0.13

1

I2 + V2

1

2

I3

3

+ V3

j0.25

–

–

–

0

0¢ V0a (a)

(b)

zF = j V0a = 1 + j

Short Circuit Studies

IF

IF

=

=

V0a Z11 + zF

=

=

1 j 0.2274

= - j 4.3975 p.u.

V0a Z 22 + zF

IF

=

1 j (0.1274 + 0.1)

1 = - j 4.2644 p.u. j 0.2345

=

V0a Z 33 + zF

=

1 = - j 4.5146 p.u. j 0.2215

V

F

= zFIF

V2( F ) = V0a - Z 21 I F = 1 - j 0.1061 (- j 4.3975) = 0.53342 p.u. and V3( F) = V0a - Z 31 I F = 1 - j 0.0981 (- j 4.3975) = 0.5686 p.u.

I12( F )

= (V1( F ) - V2( F ) )

1 1 = (0.43975 - 0.53342) j 0.08 j 0.08

= - j 1.1709 p.u. and I

F

=I

F

V j

F

V

F

1 j 0.06

1 j 0.06

5.13

5.14

Computer Techniques in Power System Analysis

I0¢3 F

V0a

V

F

1 j 0.02

I0¢

F

V0a

V

F

1 j 0.25

I

F

V

F

V

1 j 0.02 1 j 0.25

1 j 0.13

F

j

z a,b,c 1

2

1 5

4

j 1 j 0.13

¥ 1

j

3

2 4

2

3

3 (b)

(a) a

a¢

b

b¢

c

c¢ (c)

4

Short Circuit Studies

5.15

T

È- U Í- U Î

-U

-U

-U

-U

¥

where U

ÈV1a ,b,c ˘ Í a ,b,c ˙ ÍV2 ˙ Í ˙ U 0 ˘ ÍV3a ,b,c ˙ =0 ˙ 0 U ˙˚ Í Í ˙ ÍV4a ,b,c ˙ Í a ,bb,c ˙ ÍÎV5 ˙˚ Vi a ,b,c Via , Vib , and Vic , of element i i = 1, 2,

ÈV1 ˘ ÍV ˙ Í 2˙ 1 0˘ ÍV3 ˙ Í ˙=0 0 1 ˙˚ Í ˙ ÍV4 ˙ Í ˙ ÎV5 ˚

È-1 -1 -1 Í-1 -1 -1 Î

ÈU 0 0 Í0 U 0 Í ÍÎ 0 0 U

and

a ,b , c ÈV1a ,b,c ˘ È z11 Í a ,b , c ˙ Í ÍV2 ˙ Í 0 Í a ,b , c ˙ Í ÍV3 ˙ = Í 0 ÍV a ,b,c ˙ Í 0 Í 4 ˙ Í ÍÎV a ,b,c ˙˚ ÍÎ 0 5

0

È I1a ,b,c ˘ Í a ,b,c ˙ Í I2 ˙ U U ˘ Í a ,b,c ˙ I U U˙ Í 3 ˙ = 0 ˙Í ˙ U U ˙˚ Í ˙ Í I 4a ,b,c ˙ Í a ,b,c ˙ ÍÎ I 5 ˙˚ 0

0

a ,b , c z22

0

0

0

a ,b , c z33

0

0

0

0

0

a ,b , c z44 a ,b , c z54

0 ˘ È I1a ,b,c ˘ ˙Í ˙ 0 ˙ Í I 2a ,b,c ˙ ˙Í ˙ 0 ˙ Í I 3a ,b,c ˙ a ,b , c ˙ Í a ,b , c ˙ z45 I ˙Í 4 ˙ a ,b , c ˙ Í a ,b , c ˙ z55 ˚ Î I5 ˚ a ,b , c z45 and

a ,b , c z54 , where

5.16

Computer Techniques in Power System Analysis a ,b , c a ,b , c ˘˚ z45 = ÈÎ z54

T

n n

a ,b ,c a , b , c a ,b ,c Vbus = Z bus I bus a ,b ,c I bus

or where

a ,b ,c Vbus

a ,b ,c a , b , c = Ybus Vbus

ÈV1a ˘ Í b˙ ÍV1 ˙ Í c˙ ÍV1 ˙ Í ˙ Í ˙ ÍV2a ˙ a ,b ,c ˘ Í b ˙ ÈV1 ˙ Í V Í 2˙ a ,b ,c ÍV2 ˙ a ,b ,c Í ˙ c = V2 = Í ˙ ; I bus Í ˙ ˙ Í ˙ Í Í ˙ ÍÎVna ,b,c ˙˚ Í ˙ Í ˙ Í ˙ ÍVna ˙ Í b˙ ÍVn ˙ Í c˙ ÎVn ˚

È I1a ˘ Í b˙ Í I1 ˙ Í c˙ Í I1 ˙ Í ˙ Í ˙ Í I 2a ˙ a ,b ,c Í b ˙ È I1 ˘ ˙ Í I Í 2˙ a ,b ,c Í I2 ˙ Í ˙ c = I2 = Í ˙ Í ˙ ˙ Í ˙ Í Í ˙ ÍÎ I na ,b,c ˙˚ Í ˙ Í ˙ Í ˙ Í I na ˙ Í b˙ Í In ˙ Í c˙ Î In ˚

and

a ,b , c Ybus

a ,b ,c È Z11 Í a ,b ,c Z bus =Í Í Z a,b,c Î n1

n

a ,b ,c Z12

Z na2,b,c

Z1an,b,c ˘ ˙ ˙ a ,b ,c ˙ Z nn ˚

Short Circuit Studies

5.17

a b c

1

a

Three-Phase Transmission Network

b

m

c a b

n

c

a ,b ,c Z bus

a ,b ,c a ,b ,c a ,b ,c ÈÎVbus ˘˚ = ÈÎ Z bus ˘˚ ÈÎ I bus ˘˚ + [ B ] ÈÎV0a ,b,c ˘˚

where ÈU ˘ ÍU ˙ [ B ] = Í ˙ and V0a,b,c Í ˙ Í ˙ ÎU ˚

ÈV0a ˘ Í ˙ = ÍV0b ˙ Í c˙ ÍÎV0 ˙˚

n port description in the phase impedance form that can be used

VFa ,b,c

= Z Fa,b,c I Fa,b,c pth bus

I Fa,b,c

-1

= ÈÎ Z Fa,b,c + Z app,b,c ˘˚ V0a,b,c

I ap,(bF,c) = - I Fa,b,c ; I ia(F,b),c = 0

i = 1, 2,… , n iπ p

Vi a(F,b),c

= V0a ,b,c

a,b,c a ,b,c - Zip IF

i = 1, 2,… , n iπ p

5.18

Computer Techniques in Power System Analysis a ,b ,c V p( F)

= Z Fa,b,c I Fa,b,c

I aF,b,c = Y aF,b,c V aF,b,c For a fault at the pth -1

I aF,b,c = Y aF,b,c ÈÎU + Z app,b,cY aF,b,c ˘˚ V0a ,b,c I ap,(bF,c) = - I aF,b,c ; I ia(,Fb,)c = 0 i = 1, 2, … n iπ p a ,b ,c a ,b ,c V ia(,Fb,)c = V0a ,b,c - Z ip IF

i = 1, 2, … n iπ p

-1

V ap,(bF,c) = ÈÎU + Z app,b,cY aF,b,c ˘˚ V0a ,b,c Y aF,b,c is substituted Y a,b,c F

a

b

c

+ za

zb

Ia

a

+

zc

Ib

b

Ic

c +

Va Vb zg

Vc

– – –

0

0 (a)

(b)

Short Circuit Studies

ÈV a ˘ È za + z g Í b˙ Í ÍV ˙ = Í z g Í c˙ Í ÍÎV ˙˚ Î z g

5.19

zg ˘ È I a ˘

zg zb + z g zg

˙Í

˙

zg ˙ Í I b ˙ Í ˙ zc + z g ˙˚ ÍÎ I c ˙˚

a,b,c V aF,b,c = Z a,b,c F IF

È I a ˘ È ya 0 Í b˙ Í yb ÍI ˙ Í 0 Í c˙ = Í 0 0 ÍI ˙ Í y yb ÍÎ 0 ˙˚ Î a which after reduction becomes È ya ( yb + yc + y g ) ÈI a ˘ Í b˙ 1 Í - ya yb ÍI ˙ = Í Í c˙ y Í - ya yc ÍÎ I ˙˚ ÍÎ

0 0 yc - yc

˘ ÈV ˘ ˙ ÍV b ˙ ˙Í ˙ - yc ˙ ÍV c ˙ ˙Í ˙ ya + yb + yc + y g ˚ ÍV n ˙ Î ˚ a

- ya - yb

˘ ÈV a ˘ ˙ Í b˙ - yb yc ˙ ÍV ˙ ˙ Í c˙ yc ( ya + yb + y g )˙˚ ÍÎV ˚˙

- ya yb

- ya yc

yb ( ya + yc + y g ) - yb yc a,b,c I a,b,c = Y a,b,c F F VF

where za, zb, zc, and zg are the impedances, and ya, yb, yc, and yg admittances, and y = ya + yb + yc + yg Z Fa,b,c c

YFa,b,c yb = yc

YFa,b,c

È ya y g Íy + y g Í a =Í 0 Í Í 0 Í Î

˘ 0 0˙ ˙ 0 0˙ ˙ 0 0˙ ˙ ˚

Z Fa,b,c become a, b,

5.20

Computer Techniques in Power System Analysis

ya

=

YFa,b,c

a

b

È0 1Í Í0 yÍ ÍÎ0

c

0

yb ( yc + y g ) - yb yc a

b

c

yb

ya

yg

˘ ˙ - yb yc ˙ with y = yb + yc + yg ˙ yc ( yb + y g )˙˚

0

a

yc

b yb

c yc

a ya

b yb

yc

yg 0

0

0

0

(a)

(b)

(c)

(d)

ya = yg YFa,b,c

=

0 È0 1 Í0 y y b c yb + yc Í ÍÎ0 - yb yc

0 ˘ - yb yc ˙ ˙ yb yc ˙˚ yg

YFa,b,c

=

È ya ( yb + yc ) 1Í - ya yb yÍ ÍÎ - ya yc

- ya yb yb ( ya + yc ) - yb yc

- ya yc ˘ - yb yc ˙ ˙ yc ( ya + yb )˙˚

with y = ya + yb + yc ya = yb = yc = yF YFa,b,c

=

È 2 -1 -1˘ yF Í -1 2 -1˙ ˙ 3 Í ÍÎ- 1 -1 2˙˚

c

Three-Phase Components Type of Fault

Z Fa,b,c a

b

c

zF

zF

YFa,b,c

zF

È zF + z g Í Í zg Í Î zg

zg

zg zF + z g zg

˘ ˙ zg ˙ zF + z g ˙˚

zg

È yt + 2 yF 1Í yt - yF 3Í ÍÎ yt - yF

yt - yF yt + 2 yF yt - yF where yt =

yt - yF ˘ yt - yF ˙ ˙ yt + 2 yF ˙˚ 1 zF + 3z g

Three-Phase-to-Ground a

b

c

zF

zF

Three-Phase a zF

c

È zF Í0 Í ÍÎ 0

0 0˘ • 0˙ ˙ 0 • ˙˚

È yF Í0 Í ÍÎ 0

0 0˘ 0 0˙ ˙ 0 0˙˚

5.21

Line-to-Ground

b

Short Circuit Studies

È 2 -1 -1˘ yF Í -1 2 -1˙ ˙ 3 Í ÍÎ-1 -1 2˙˚

zF

5.22

Three-Phase Components Type of Fault

a

b zF

c zF

È• Í Í0 Í Î0

zF

Line-Line-to-Ground a

b b

zF

0 zF + z g zg

˘ ˙ zg ˙ ˙ zF + z g ˚ 0

YFa,b,c

È Í Í0 Í Í0 Í Í Í Í0 ÍÎ

0 zF + z g z F2

+ 2 zF z g - zF

z F2

+ 2 zF z g

c zF

Line-to-Line

È0 0 0 ˘ yF Í 0 1 -1˙ ˙ 2 Í ÍÎ0 -1 1˙˚

˘ ˙ ˙ 0 ˙ - zg ˙ z F2 + 2 z F z g ˙ ˙ zF + z g ˙ ˙ z F2 + 2 z F z g ˙˚

Computer Techniques in Power System Analysis

Z Fa,b,c

Short Circuit Studies

5.23

a,b,c Z bus ,

a ,b , c a ,b , c z45 and z54

V4a ,b,c and V5a ,b,c , and currents I 4a ,b,c and I 5a ,b,c ÈV4a ,b,c ˘ Í a ,b , c ˙ ÍÎV5 ˙˚

=

a ,b , c È z44 Í a ,b , c ÍÎ z54

a ,b , c ˘ È I 4a ,b,c ˘ z45 ˙ Í a ,b , c ˙ a ,b , c z55 ˙˚ ÍÎ I 5 ˙˚

a ,b , c where, z45

aa È z45 Í ba = Í z45 Í ca ÍÎ z45

ab z45 bb z45 cb z45

ac ˘ z45 ˙ a ,b , c T bc z45 ˙ = ÈÎ z54 ˘˚ cc ˙ z45 ˙˚

a ,b , c z45

ÈV40,1, 2 ˘ È S 3 Í 0,1, 2 ˙ = Í ÍÎV5 ˙˚ Î 0

a ,b , c 0 ˘ È z44 Í a ,b , c S3 ˙˚ ÍÎ z54

È S3 z a ,b,c S3-1 = Í 44 a ,b ,c -1 S3 ÍÎ S3 z54

a ,b , c ˘ È S3-1 z45 ˙Í a ,b , c z55 ˙˚ ÍÎ 0

0 ˘ È I 40,1, 2 ˘ ˙ ˙Í S3-1 ˙˚ ÍÎ I 50,1, 2 ˙˚

0,1, 2 a ,b ,c -1 S3 z45 S3 ˘ È I 40,1, 2 ˘ È z44 = ˙ Í ˙ Í 0,1, 2 a ,b ,c -1 S3 z55 S3 ˙˚ ÍÎ I 50,1, 2 ˙˚ ÍÎ z54

0,1, 2 ˘ È I 40,1, 2 ˘ z45 ˙Í ˙ z505,1, 2 ˙˚ ÍÎ I 50,1, 2 ˙˚

S 0,1, 2 0,1, 2 complete transposition is considered, then z44 and z55 0,1, 2 z45 can be obtained as

0,1, 2 z45

=

a ,b ,c -1 S3 z45 S3

È1 1 1Í = Í1 a 3 Í1 a 2 Î

aa 1 ˘ È z45 Í ˙ ba a 2 ˙ Í z45 Í ca a ˙˚ ÍÎ z45

ab z45 bb z45 cb z45

z4ac5 ˘ È1 1 ˙ bc Í 2 z45 ˙ Í1 a cc ˙ Í z45 ˙˚ Î1 a

1˘ ˙ a˙ a 2 ˙˚

5.24

Computer Techniques in Power System Analysis 0,1, 2 z45 as

(

(

0,1, 2 z45

( z45aa + az45bb + a 2 z45cc ) ˘˙ ) ( ) ba ca + ( z45 + z45 ) + ˙˙ bc ba ca ac z z a z z + + + + ) ( 45 45 ) ( 45 45 ) a ( z ab + z cb ) ˙ 45 45 2 ab cb ˙ + a z + z ) ( 45 45 ) ac bc + a 2 ( z45 + z45 ) ˙˙

È Í z aa + z bb + z cc 45 45 Í 45 ab ca Í + z + z + z bc 45 45 45 Í Í + z ba + z ac + z cb 45 45 45 Í Í Í Í aa bb 2 cc Í z45 + az45 + a z45 Í ab ac Í+ z455 + z45 1Í ba bc = Í+ a z45 + z45 3Í 2 ca cc Í+ a z45 + z45 Í Í Í z aa + a 2 z bb + az cc 45 45 Í 45 Í+ z ab + z ac 45 45 Í Í+ a z ca + z cb 45 45 Í Í+ a 2 z ba + z bc 45 45 Í Í Î

(

( (

(

(

(

(

)

(

)

(

)

)

aa z45

)

)

cc + az45

( z45aa + z45bb + z45cc ) ac ba cb + a ( z45 + z45 + z45 ) 2 ab bc ca + a ( z45 + z45 + z45 )

)

)

bb + a 2 z45

(

aa z45

bb + a 2 z45 5

( + )+ ab ba + z45 a ( z45 ) 2 ac ca + a ( z45 + z45 ) +

bc z45

aa bb cc ab ac ba bc ca cb m z45 = z45 = z45 = z45 = z45 = z45 = z45 = z45 = z45 = z45 0,1, 2 z45 .

a ,b , c z45 ,

z

0,1, 2

)

cb z45

( z45aa + az45bb + a 2 z45cc ) ( z45aa + z45bb + z45cc ) bc cb + ( z45 + z45 ) + a ( z45ac + z45ca ) +a ( z45ab + z45bc + z45ca ) ab ba ac ba cb + a 2 ( z45 + z45 + a 2 ( z45 + z45 + z45 ) )

a ,b , c z45

0,1, 2 È z11 Í Í 0 Í =Í 0 Í 0 Í ÍÎ 0

cc + az45

0

0

0,1, 2 z22

0

0

0

0

0,1, 2 z33

0

0

0

0,1, 2 z44

0

0

0,1, 2 z54

0 ˘ ˙ 0 ˙ ˙ 0 ˙ 0,1, 2 ˙ z45 ˙ 0,1, 2 ˙ z55 ˚

˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˙ ˚

Short Circuit Studies

5.25

where 0,1, 2 zkk

È zk s + 2 zk m =Í 0 Í ÍÎ 0

zk s

0 - zk m 0

m È3 z45 ˘ ˙ ; z 0,1, 2 = Í 0 45 Í ˙ Í 0 - zk m ˙˚ Î

0 0˘ ˙ 0 0˙ 0 0˙˚

0 0

zk s

zks is the self inductance of each phase and zkm is the mutual inductance between m k z45

V0a = Va –0 , V0b = Va – - 120 , V0c = Va – - 240 . V S = V00,1, 2

1 ˘ ÈV0a ˘ È1 1 Í ˙ 1 ˙ Í a 2 ˙ ÍV0b ˙ = Í1 a 3Í Í c˙ 2 ˙ a ˚ ÍÎV0 ˙˚ Î1 a

È1 1 1 Í = Í1 a 3Í 2 Î1 a

1 ˘ È Va –0∞ ˘ È 0 ˘ Í ˙ ˙ a 2 ˙ Í Va – - 120∞ ˙ = Í 3 ˙ Va Í ˙ a ˙˚ ÍÎ Va –120∞ ˙˚ ÎÍ 0 ˙˚

n components, the Z ÈÎV admittance form as Y

0,1, 2 bus ˘ ˚

0,1, 2 F

Z

,1, 2 ,1, 2 ˘˚ ÈÎ I 0bus ˘˚ + [ B ] ÈÎV = ÈÎ Z 0bus 0,1, 2 F

and Y

0,1, 2 ˘˚ 0

Z 0F,1, 2 , or in the

0,1, 2 F

Z Fa,b,c

or YFa,b,c YF0,1, 2 = S3YFa,b,c S3-1 È1 1 È1 1 1 ˘ È yF 0 0˘ 1 Í 1 Í 2˙ Í 2 ˙ = Í1 a Í1 a a ˙ Í 0 0 0˙ 3Í 3 2 Í1 a a ˙˚ ÎÍ 0 0 0˙˚ Î Î1 a È1 1 1 ˘ È yF yF yF ˘ 1Í ˙ = Í1 a a 2 ˙ Í 0 0 0 ˙ Í ˙ 3 Í1 a 2 a ˙ ÍÎ 0 0 0 ˙˚ Î ˚ È yF yF yF ˘ 1Í = Í yF yF yF ˙˙ 3 ÎÍ yF yF yF ˙˚

z F0,1, 2

=

S3 z Fa,b,c S3-1

È zF + 3z g Í 0 =Í Í 0 Î

0 zF 0

0˘ ˙ 0˙ z F ˙˚

1˘ ˙ a˙ a 2 ˙˚

5.26

Computer Techniques in Power System Analysis

Symmetrical Components Type of Fault a

b

c

zF

zF

Y 0F,1, 2

Z 0,1,2 F

zF

È zF + 3z g Í 0 Í Í 0 Î

zg

0 zF 0

È yF Í0 Í ÍÎ 0

0˘ ˙ 0˙ z F ˙˚

0 yF 0

where y0 =

0˘ 0˙ ˙ yF ˙˚ 1 zF + 3z g

Three-Phase-to-Ground a

b

c

zF

zF

È• Í0 Í ÍÎ 0

zF

0 zF 0

0˘ 0˙ ˙ z F ˙˚

È0 0 0 ˘ y F Í0 1 0 ˙ Í ˙ ÍÎ0 0 1 ˙˚

Three-Phase a

b

c

È1 1 1˘ yF Í 1 1 1˙ ˙ 3 Í ÍÎ1 1 1˙˚

zF

Line-to-Ground a

b zF

c zF - zF - zF È 2 zF ˘ Í- z 2 z + 3z - z + 3z ˙ ( ) F F g F g ˙ 3 ( z F2 + 2 z F z g ) Í ÍÎ - z F - ( z F + 3 z g ) 2 z F + 3 z g ˙˚ 1

zg

Line-Line-to-Ground a

b zF

c zF

Line-to-Line

È0 0 0 ˘ yF Í 0 1 -1˙ ˙ 2 Í ÍÎ0 -1 1˙˚

Short Circuit Studies

5.27

Z 0F,1, 2 1

I F0,1, 2 = ÈÎ Z 0F,1, 2 + Z 0pp,1, 2 ˘˚ V00,1, 2 0,1, 2 0,1, 2 Vi0( F,1,)2 = V00,1, 2 - Z ip IF

i = 1, 2, … , n iπ p

0,1, 2 0,1, 2 0,1, 2 V p( F) = Z F I F

Y 0F,1, 2 -1

I F0,1, 2 = Y 0F,1, 2 ÈÎU + Z 0F,1, 2Y 0F,1, 2 ˘˚ V00,1, 2 and 0,1, 2 0,1, 2 Vi0( F,1,)2 = V00,1, 2 - Z ip IF

i = 1, 2, … , n iπ p -1

V p0(,1F, 2) = ÈÎU + Z 0pp,1, 2Y 0F,1, 2 ˘˚ V00,1,2

Z 0F,1, 2

I F0,1, 2 = ÈÎ Z F0,1, 2 +

-1 Z 0pp,1, 2 ˘˚ V00,1, 2

È z F + 3 z g + Z (pp0) Í 0 =Í Í ÍÎ 0

È È I F(0) ˘ Í Í (1) ˙ Í ÍIF ˙ = Í Í (2) ˙ Í z F ÍÎ I F ˙˚ Í Î

0 z F + Z (p1p)

˘ ˙ 3 ˙ V 1) ˙ a + Z (pp ˙ ˙˚ 0 0

0

˘È 0 ˘ ˙Í ˙ ˙ Í 3 ˙ Va 0 ˙Í ˙ z F + Z (pp2) ˙˚ Î 0 ˚ 0

5.28

Computer Techniques in Power System Analysis

È I F(a ) ˘ È1 1 Í (b ) ˙ 1 Í 2 -1 0,1,2 Í I F ˙ = S3 I F = Í1 a 3 Í (c ) ˙ Í1 a Î ÍÎ I F ˙˚

1˘ ˙ a ˙ I F0,1,2 a 2 ˙˚

È ÈVF(0) ˘ Í Í (1) ˙ Í ÍVF ˙ = Í Í (2) ˙ Í z F ÍÎVF ˙˚ Í Î

ÈVF(a ) ˘ Í (b ) ˙ -1 0,1,2 ÍVF ˙ = S3 VF Í (c ) ˙ ÍÎVF ˙˚

È Í Í zF Í =Í Í zF Í Í Í Î zF

È Í Í zF Í =Í Í zF Í Í Í Î zF

1 1) + Z (pp

1 1) + Z (pp

1 1) + Z (pp

˘ ˙ ˙ ˙ – - 120 ˙ Va ˙ ˙ ˙ – - 240 ˙ ˚

˘ ˙ 3zF ˙ V 1) ˙ a + Z (pp ˙ ˙˚ 0 0

zF 1) + Z (pp

zF + Z (p1p) 1 1) + Z (pp

ÈVi(0) ˘ È0˘ È Z ip(0) (F ) Í ˙ Í ˙ Í ÍVi(1) ˙ Í 0 = V 3 a (F ) Í ˙ ÍÍ ˙˙ Í 0˚ ÍVi(2) ˙ Í 0 Î Î Î (F ) ˚

˘ ˙ ˙ ˙ – - 120 ˙ Va ˙ ˙ ˙ – - 240 ˙ ˚

0 Z ip(1) 0

0 ˘È 0 ˘ ˙Í ˙ 0 ˙ Í I F(1) ˙ ˙Í ˙ Z ip( 2) ˙˚ Î ˚

Ê Z ip(1) ˆ Vi(1) = 3 1 Va i = 1, 2, … , n Á (F ) (1) ˜ Ë z F + Z pp ¯ iπ p

V1aF,b,c = S3-1ViF0,1, 2 for i = 1, 2, … , n iπ p È1 1 1 Í 2 = Í1 a 3Í Î1 a

1˘È 0 ˘ Ê Z ip(1) ˆ ˙ Í (1) ˙ a ˙ ÍVi (F) ˙ Va = Á1 (1) ˜ Ë z F + Z pp ¯ ˙ 2˙ Í a ˚Î 0 ˚

1 È Í Í – - 120 Í– - 240 Î

˘ ˙ ˙ Va ˙ ˚

Short Circuit Studies

5.29

a YF0,1, 2 is used because for this type of fault, Z F0,1, 2 YF0,1, 2

I F0,1, 2

=

È1 1 1˘ yF Í 1 1 1˙ ˙ 3 Í ÍÎ1 1 1˙˚

Ï È Z (0) È1 1 1˘ Ô È1 0 0˘ Í pp y Ô = F Í1 1 1˙ Ì Í0 0 0˙ + Í 0 Í ˙ Í ˙ Í 3 ÍÎ1 1 1˙˚ Ô ÍÎ0 0 1 ˙˚ Í 0 ÔÓ Î

With this assumption, Z (1) pp

I F0,1, 2

È1 1 yF Í = 1 1 3 Í ÍÎ1 1

=

Z (0) pp + 3 z F

-1

0 1) Z (pp

0

0 ˘ 1 1 1˘ ¸ È 0 ˘ ˙ y È ÔÔ Í ˙ 0 ˙ ¥ F Í1 1 1˙ ˝ Í 3 ˙ Va ˙ ˙ 3 Í ÍÎ1 1 1˙˚ Ô ÍÎ 0 ˙˚ Z (pp2) ˙˚ Ô˛

= Z (2) pp

yF È (0) y F Z (0) pp Í1 + Z pp 3 3 1˘ Í y yF F 1˙ Í Z (1) 1 + Z (1) pp ˙ Í pp 3 3 1˙˚ Í y y (2) (2) F F Í Z Z pp ÍÎ pp 3 3 È1˘ 3 Í1˙ V a (2) Í ˙ + Z (1) pp + Z pp ÍÎ1˙˚

yF 3 y F Z (1) pp 3 yF 1 + Z (2) pp 3 Z (0) pp

˘ ˙ ˙ ˙ ˙ ˙ ˙ ˙˚

-1

È0˘ Í ˙ Í 3˙ Í0˙ Î ˚

Va

be used

(U + B ) 1 = U uvT, where u and v

if B

v Tu = a

B

(1 + a ) a =  diag (bii ) = v T u i B ˆ Ê U+B ÁË U - (1 + a ) ˜¯ and

5.30

Computer Techniques in Power System Analysis

3 ˘ È I ((Fa )) ˘ È 1) ( 2) ˙ Í (b ) ˙ Í Z (pp0) + Z (pp + Z pp + 3 z F ˙ Va Í I (F ) ˙ = Í 0 Í ˙ Í (c ) ˙ ˙˚ 0 Î I (F ) ˚ ÍÎ ÈV((F0)) ˘ È0˘ Í (1) ˙ 0,1, 2 0,1, 2 -1 Í 3 ˙ Z ( 2) ÍV( F ) ˙ = ÈÎU + Z pp YF ˘˚ Í ˙ pp Í ( 2) ˙ Î0˚ ÍÎV( F ) ˙˚ È ˘ - Z (pp0) Í (0) ˙ 3 ( 2) = (0) Í Z pp + Z pp + 3 z F ˙ Va (1) ( 2) Z pp + Z pp + Z pp + 3 z F Í ˙ - Z (pp2) Î ˚ 1) Z (pp0) + Z (pp + Z (pp2) + 3 zF = Z1g

˘ È 3zF Í ˙ Z1g ˙ ÈV((Fa)) ˘ Í ( 0) ( 2) ˙ 2 (1) Í Z pp + a Z pp + aZ pp Í (b ) ˙ 2 ˙ Va ÍV(F ) ˙ = Ía Z1g ˙ Í (c ) ˙ Í ( 0) (1) 2 ( 2) ˙ ÎV(F ) ˚ Í Z pp + aZ pp + a Z pp ˙ Í Í a˙ Z1g Î ˚ È Z ip(0) ˘ Í ˙ Í Z1g ˙ (0) ÈVi (F ) ˘ È 0 ˘ Í Z (1) ˙ Í (1) ˙ Í ˙ Í ip ˙ V i = 1, 2, … , n = V V 3 3 Í i (F ) ˙ Í ˙ a Í Z1g ˙ a i π p Í (2) ˙ Î 0 ˚ Í ( 2) ˙ V Î i (F ) ˚ Í Z ip ˙ Í ˙ ÍÎ Z1g ˙˚ È Z ip(0) ˘ Í ˙ Z1g ˙ Í ÍÊ Z (1) ˆ ˙ i = 1, 2, … , n ip ˙ V = 3 ÍÁ1 ÍË Z1g ˜¯ ˙ a i π p Í ˙ Í Z ip( 2) ˙ Í ˙ Z1g ˙˚ ÍÎ

Short Circuit Studies

È Z ip(0) + Z ip(1) + Z ip( 2) ˘ 1 Í ˙ Z1g Í ˙ ÈVia(F ) ˘ Í ˙ ( 0 ) ( 1 ) ( 2 ) 2 Í ˙ Í Z ip + a Z ip + aZ ip ˙ i = 1, 2, … , n b 2 ÍVi (F ) ˙ = Ía Va ˙ iπ p Z1g Í ˙ Í ˙ ÍÎVic(F ) ˙˚ Í ( 0) (1) 2 ( 2) ˙ Í a - Z ip + aZ ip + a Z ip ˙ Í ˙ Z1g Î ˚

1

2

3

1

1

2

3 2

5

6 4

Line Reactances Bus code

Self-impedance +ve seq.

– ve seq.

Mutual impedance

Zero seq

Zero seq

—

—

—

—

Generator Reactances Generator

Impedance +ve seq.

1 2

V0a = 1 + j 0

-ve seq.

Coupling element

Zero seq

5.31

5.32

Computer Techniques in Power System Analysis

0 1 1 2 0 0,1, 2 Z bus =2 1 2 0 3 1 2

1 0 1 2 0 . 0291 0 0 È Í 0 0.124 0 Í 0 0.07905 Í 0 Í Í Í 0.0198 0 0 Í 0 0 . 118 0 Í Í 0 0 0.07262 Í Í Í 0.0054 0 0 Í Í 0 0.101 0 Í 0 0.0556 Î 0

0 0.0198 0 0

2 1

2 0

0

0.118 0 0 0.07262

0.0639 0 0 0 0.1316 0 0 0 0.08776 0.0105 0 0

0

0

0.1065 0 0 0.0 06442

3 0 1 0.0054 0

˘ 0.101 0 ˙ ˙ 0 0.0556 ˙ ˙ ˙ 0.0105 0 0 ˙ ˙ 0 0.1065 0 ˙ 0 0 0.06442˙ ˙ ˙ 0.017 0 0 ˙ ˙ 0 0.1188 0 ˙ ˙ 0 0 0.0745 ˚

(A)

I F0,1, 2

0 È Í 3 = Í (1) ÍZ + z F Í 33 ÍÎ 0

I Fa,b,c

=

V3(0,F1,)2

˘ È 0 ˘ ˙ Í ˙ È 0 ˘ 3 ˙ Í ˙=Í = 14.58˙ ˙ ˙ Í 0.1188 ˙ Í ˙ Í ˙ ÍÎ 0 ˙˚ 0 ˙˚ Î ˚

S3-1 I F0,1,2

È Í =Í Íz Í F ÍÎ

È1˘ Í ˙ = 8.417 Ía 2 ˙ Ía˙ Î ˚

˘ ˙ È0 ˘ 3zF ˙ Í ˙ = 0 (1) ˙ Í ˙ + Z 33 ˙ ÍÎ0˙˚ ˙˚ 0 0

and V3a,b,c F

=

S3-1V3(0,1,2 F)

È0 ˘ = Í0 ˙ Í ˙ ÍÎ0˙˚

2

0 0

0

Short Circuit Studies

0 ˘ È Í ˙ È 0 ˘ (1) Ê ˆ Z 0,1, 2 13 Í ˙ = Í0.259˙ V1(F ) = 3 Á1 ˙ Í Ë z + Z (1) ˜¯ ˙ Í F 33 Í ˙ ÍÎ 0 ˙˚ ÍÎ ˙˚ 0 0 ˘ È Í ˙ È 0 ˘ (1) Ê Z 23 ˆ ˙ Í V2(0,F1,)2 = Í 3 Á1 = 0.1793˙ ˙ Í Ë z + Z (1) ˜¯ ˙ Í F 33 Í ˙ ÍÎ 0 ˙˚ ÍÎ ˙˚ 0

=

V1(a,b,c F)

S3-1V1(0,1,2 F)

-1 0,1,2 V2(a,b,c F ) = S3 V2(F )

0,1, 2 I 23( F)

0 ˘ ˘ È È 0 ˘ ˙ Í 0.1793 ˙ Í ˙ ˙ = Í 0.06 ˙ = Í2.9883˙ Í ˙ ˙ ˚ ÍÎ 0 ˙˚ ÍÎ 0 ˙˚ 0 ˘ 0 ˘ È È 0 ˘ Í 0.259 ˙ Í ˙ ˙ = 2.59˙ V1((1F) ) - V3((1F) ) ˙ = Í Í ˙ ˙ Í 0.1 ˙ ÍÎ 0 ˙˚ 0 ˚ ÍÎ 0 ˙˚

0 È Í (1) (1) = Í y23 V2(F ) - V3((1F) ) Í 0 Î

(

È Í (1) 0,1, 2 I13( = F) Í y13 Í Î

È1˘ Í 2˙ = 0.1485 Ía ˙ Ía˙ Î ˚ È1˘ Í 2˙ = 0.1035 Ía ˙ Ía˙ Î ˚

)

(

)

0,1, 2 0,1, 2 I12(1)( F ) = I12(2)(F )

and a,b,c a,b,c I12(1)( F ) = I12(2)(F )

0,1, 2 I12(1)( F)

0 È Í (1) (1) = Í y12 V1(F ) - V2((1)F ) Í 0 Î

(

)

0 ˘ ˘ È È 0 ˘ ˙ Í 0.0797 ˙ Í ˙ ˙ = Í 0.05 ˙ = Í1.594˙ Í ˙ ˙ ˚ ÍÎ 0 ˙˚ ÍÎ 0 ˙˚

5.33

5.34

Computer Techniques in Power System Analysis

a,b,c I 23( F)

a,b,c I13( F)

a,b,c I12(1)( F)

=

0,1,2 S3-1 I 23( F)

È1˘ Í 2˙ = 1.725 Ía ˙ Ía˙ Î ˚

=

0,1,2 S3-1 I13( F)

È1˘ Í ˙ = 1.495 Ía 2 ˙ Ía˙ Î ˚

=

0,1,2 S3-1 I12( F)

È1˘ Í ˙ = 0.9203 Ía 2 ˙ Ía˙ Î ˚

(B)

I (0F,1), 2

È1˘ 3 Í1˙ = = (0) (1) ( 2) Í ˙ 0.017 + 0.1188 + 0.0745 Z 33 + Z 33 + Z33 + 3 z F ÍÎ1˙˚ 3

I (a,b,c F)

V(0F,1), 2

=

S3-1 I (0,1,2 F)

È1˘ È8.236˘ Í1˙ = Í8.236˙ Í˙ Í ˙ ÍÎ1˙˚ ÍÎ8.236˙˚

È14.265˘ =Í 0 ˙ Í ˙ ÍÎ 0 ˙˚

( 0) È ˘ - Z 33 3 Í ( 0) ˙ ( 2) 0,1, 2 = V3(F ) = (0) Í Z 33 + Z 33 + 3 z F ˙ (1) ( 2) Z 33 + Z 33 + Z 33 + 3 z F Í ( 2) ˙ - Z 33 Î ˚ -0.0170 È ˘ 3 Í0.0170 + 0.0745˙ = 0.017 + 0.1188 + 0.0745 Í ˙˚ -0.0745 Î È-0.1400˘ = Í 0.7536˙ Í-0.6136˙ Î ˚

V1(0F,1,)2

È0˘ 3 Í ˙ = Í 3 ˙ - (0) (1) ( 2) Í 0 ˙ Z 33 + Z 33 + Z 33 + 3 z F Î ˚

(0) È Z13 ˘ Í (1) ˙ Í Z13 ˙ Í ( 2) ˙ ÍÎ Z13 ˙˚

Short Circuit Studies

È0˘ 3 Í ˙ = Í 3˙ 0.017 + 0.1188 + 0.0745 Í0˙ Î ˚ V2(0,F1,)2

È0˘ 3 Í ˙ = Í 3 ˙ - ( 0) (1) ( 2) Í 0 ˙ Z 33 + Z 33 + Z 33 + 3 z F Î ˚

È0.0054˘ È-0.0444˘ Í 0.101 ˙ = Í 0.9002 ˙ ˙ Í ˙ Í ÍÎ0.0556˙˚ ÍÎ-0.4579˙˚ ( 0) È Z 23 ˘ Í (1) ˙ Í Z 23 ˙ Í ( 2) ˙ ÍÎ Z 23 ˙˚

È0˘ 3 Í ˙ = Í 3˙ 0.0170 + 0.1188 + 0.0745 Í0˙ Î ˚

V3(a,b,c F)

V1(a,b,c F)

V2(a,b,c F)

0,1,2 I13( F)

( ( (

È 0.0105 ˘ È -0.0865˘ Í ˙ Í ˙ Í 0.1065 ˙ = Í 0.8549 ˙ Í0.06442˙ Í-0.5306˙ Î ˚ Î ˚

=

S3-1V3(0,1,2 F)

0 È ˘ Í = Í-0.1212 - j 0.6836˙˙ ÍÎ-0.1212 + j 0.6836˙˚

=

S3-1V1(0,1,2 F)

0.2297 È ˘ Í ˙ = Í-0.1533 - j 0.67902˙ Í-0.1533 + j 0.67902˙ Î ˚

=

S3-1V2(0,1,2 F)

0.1372 È ˘ Í ˙ = Í-0.1436 - j 0.6972˙ Í-0.1436 + j 0.6972˙ Î ˚

(0) È V1((0) F ) - V3(F ) Í (1) = Í V1(F ) - V3((1)F ) Í (2) (2) ÍÎ V1(F ) - V3(F )

) ) )

È 0.0956 ˘ Í ˙ ( 0) ˘ Í 0.15 ˙ È0.6373˘ y13 ˙ Í 0.1466 ˙ Í ˙ y1(13) ˙ = Í ˙ = Í 1.466 ˙ 0 1 . ( 2) ˙ ˙ Í ˙ Í y13 ˙˚ Í 0.1557 ˙ Î 1.557 ˚ ÍÎ 0.1 ˙˚

( ) ( y12(0()1) + ym(012) )˘˙ È0.3102˘ Í ˙ (1) 0,1,2 I12(1)( y12 ˙ = Í 0.906 ˙ ( ) F) (1) ˙ Í ( 2) ˙ y12 ( ) ˙˚ Î 1.454 ˚ (1) (0) ( 0) ( 0) È(V1((0) ˘ F ) - V2(F ) ) ( y12 ( 2 ) + ym12 ) Í ˙ È0.2216˘ (1) (1) (1) 0,1,2 Í ˙ = Í 0.906 ˙ I12(2)(F ) = (V1(F ) - V2(F ) ) y12( 2) ˙ Í ˙ Í ( 2) Í(V (1) - V (1) ) ˙ ÍÎ 1.454 ˙˚ y 2(F ) 12 ( 2 ) Î 1(F ) ˚ (0) È V1((0) F ) - V2(F ) Í (1) = Í V1((1) F ) - V2(F ) Í (2) (2) ÍÎ V1(F ) - V2(F )

5.35

5.36

Computer Techniques in Power System Analysis

0,1,2 I 23( F)

( ( (

(0) È V2(F) - V3((0)F ) Í (1) = Í V2(F) - V3((1)F ) Í (1) (1) ÍÎ V2(F) - V3(F )

) ) )

( 0) ˘ y23 È0.4458˘ (1) ˙ y23 ˙ = Í 1.688 ˙ ÍÎ1.3833 ˙˚ ( 2) ˙ y23 ˙˚

2.133 È ˘ a,b,c -1 0,1,2 I13( F ) = S3 I13(F ) = Í -0.5047 + j 0.0456 ˙ ÍÎ-0.5047 - j 0.0456˙˚ 1.542 È ˘ a,b,c -1 0,1,2 Í I12(1)( = S I = 0 . 502 + j 0.274˙ F) 3 12(1)(F ) ÍÎ- 0.502 - j0.274˙˚ 1.4905 È ˘ a,b,c -1 0,1,2 Í ˙ I12(2)( = S I = 0 . 553 + j 0 . 274 F) 3 12(2)(F ) Í- 0.553 - j 0.274˙ Î ˚ and 2.031 È ˘ a,b,c -1 0,1,2 Í ˙ I 23( F ) = S3 I 23(F ) = -0.6291 - j 0.1528 Í-0.6291 + j 0.1528˙ Î ˚

5.1 –

3

0.1 0.1 2

1 0.075 + 1–0° -

0.05 + 1–0° -

Short Circuit Studies

5.37

5.2 1

2

2

0 1 2 0 È0.05 0 0 ˘ 1 Í 0 0.10 0 ˙ Í ˙ 2 ÍÎ 0 0 0.10˙˚

Line 1 - 2 0 1 2 0 È0.1 0 0˘ Í 1 0 0.5 0 ˙ Í ˙ 2 ÍÎ 0 0 0.5˙˚

Line 1 - 3 0 1 2 0˘ 0 È0.2 0 Í 1 0 0.5 0 ˙ Í ˙ 0 0.5˙˚ 2 ÍÎ 0

0 0 È0.15 1 Í 0 Í 2 ÍÎ 0

Line 2 - 3 0 1 2 0 0 ˘ 0 È0.5 Í 0 ˙ 1 0 0.25 Í ˙ 2 ÍÎ 0 0 0.25˙˚

1 2 0 0˘ 0 0˙ ˙ 0 0˙˚ 0,1, 2, Z bus

Zbus for

Z Fa,b,c

5.3

Z a , Zb , Z c , to Z g , as 0¢ and after

Hint th

order Zbus, impose the restriction that the bus current at 0¢ is ¢

5.38

Computer Techniques in Power System Analysis

0 as the reference and then ya , yb , and yc , to obtain YFa,b,c

5.4

5.5 p

5.6 5.7 2 L-G 5.8

5.9

6

Y

Z

Y

Y

6.2

Computer Techniques in Power System Analysis

Y n I

=Y

V Pi Q i

Vi

Pi Qi Vi

qi

I1 1

+ V1

I2 + V2

2 Pi + jQi

Ii + Vi

Pgi + jQgi

i In

+ Vn

P�i + jQ�i n

Pi = Pgi - P�i Qi = Qgi - Q�i

0

PQ Vi

qi

Vi I i* = Sisp = Pi sp + jQisp Pi sp = Pgisp - P sp i

Qisp = Qgisp - Q spi

i

6.3

Power Flow Studies

(

)

(

)

Pi sp = Re Vi I i* Qisp = Im Vi I i* g

Pi sp

PV Vi

(

sp

qi

Qi

)

Pi sp = Re Vi I i* where Pi sp = Pgisp - P sp i Vi = Vi

sp

Vs sp

V-q

I R

qs

I X Pi

Qi I R

I X

swing Vs sp

slack

Ps Qs

s Ps = Pgs - P sp s Qs = Pgs - Q sps Q sps

Pgs

P sp s

Qgs

Type of Bus PQ

Pi sp Qisp

PV

Pi sp

Vq

Vs

sp

Vi

sp

q ssp

(q ssp ∫ 0)

Unknowns

Number of Buses

qi

n

Vi

Qi q i Ps

Qs

n

6.4

Computer Techniques in Power System Analysis

n +n

n

Notation Yij = i - j element of Ybus = Gij + jBij = Yij –fij Vi = ei + jfi = Vi –q i q ij = q i - q j

n

I i = Â YijV j

i = 1, 2,

,n

j =1

n

Vi I i* = Vi  Yij* V j* j=

Pi sp

Qisp

qi

Vi

Vi

iπs

Bus Mismatch Criteria Vi

Yij

P Q Bus

È n ˘ Pi sp = Re ÍVi  Yij* V j* ˙ ÍÎ j = ˙˚ n È ˘ = Re Í Vi – q i  (Gij - jBij ) V j – - q j ˙ ÍÎ ˙˚ j=

Èn ˘ = Vi ÍÂ (Gij cos q ij + Bij sin q ij ) V j ˙ ÍÎ j =1 ˙˚ Èn Qisp = Vi ÍÂ (Gij sin q ij - Bij cos q ij ) V j ÍÎ j =1

˘ ˙ ˙˚

Power Flow Studies

6.5

P V Bus Èn ˘ Pi sp = Vi ÍÂ (Gij cos q ij + Bij sin q ij ) Vi ˙ ÍÎ j =1 ˙˚ DPi

DQi

Cp

DPi £ Cp

PQ

DQi £ Cq

PQ

PV

Cq

Voltage Magnitude Criteria k

k

k DVi = Vi( k ) - Vi( k -1) DVi £ Cv

PQ

P-V PV PV s

Vs –qs

Vs Sisp = Pi sp + jQisp = Vi  Yij* V j* j =1

i = 1, 2, , n iπs

Vi

(Sisp )

*

Vi

*

n

=

i = 1, 2,

 YijV j

,n

iπs

j =1

Vi È sp 1 Í Si Vi = Yii Í Vi* Í Î

( )

*

˘ ˙ - Â YijV j ˙ j =1 ˙ i =1, 2, πi ˚ n

iπs

,n

Cv

6.6

Computer Techniques in Power System Analysis

Vs – qs

Vs n

V1( k ) =

sp sp 1 È P1 - jQ1 - Y12V2( k -1) - Y13V3( k -1) Í Y11 Î V1( k -1)*

V2( k ) =

sp sp 1 È P2 - jQ2 - Y21V1( k ) - Y23V3( k -1) Í Y22 Î V2( k -1)*

Vi( k ) =

sp sp 1 È Pi - jQi - Yi1V1( k ) Í Yii Î Vi ( k -1)*

Vn( k ) =

1 È Pnsp - jQnsp - Yn1V1( k ) Í Ynn Î Vn( k -1)*

- Y1sVs

˘ - Y1nVnVn(k -1) ˙ ˚

˘ - Y2 sVs … - Y2 nVn( k -1) ˙ ˚

- Yi (i -1)Vi(-k1) - YiiVi k -1 - YnsVs

˘ - YinVn( k -1) ˙ ˚

- YisVs

˘ - Yn ( n -1)Vn(k )( n-1) ˙ ˚

k Vi

(0)

Vi

( 0)

(i = 1, 2,

, n, i π s )

= 1–0

Vi i

k

Max

i Vi( k ) - Vi( k -1) < Cv Vi(0) = 1 – 0∞ (i π s )

.01 < Cv < .0001

P-V PV PV

Qi Qgi

Vi Qgimin < Qgi < Qgimax .

Vi PV

Qicalc Èn Qicalc = Vi ÍÂ (Gij sin q ij - Bij cos q ij ) V j ÍÎ j =1 Qgi = Qicalc + Q

i

Qgimin < Qgi < Qgimax

˘ ˙ ˙˚

6.7

Power Flow Studies

Qgi Vi Vi( k )calc =

sp 1 È Pi - jQicalc - Yi1Vi( k ) Í Yii Î Vi( k -1)*

- Yi (i -1)Vi(-k1)

˘ - YinVn( k -1) ˙ ˚

- YisVs

= ei( k ) calc + j fi( k ) calc Vi( k )calc = Vi

sp

(

q i( k )calc = tan -1 fi( k )calc ei( k )calc

)

Vi( k ) Vi( k ) new = Vi

sp

ÈÎcos q i( k ) calc + j sin q i( k ) calc ˘˚

Vi( k ) Qgi < Qgimin , set Qgi = Qgimin Qgi > Qgimax , set Qgi = Qgimax PQ Vi

( k )calc

PV PV

PV

PQ

yij = -Yij i

yij¢

j

i

j

È yij¢* ˘ * Sij = Vi I ij* = Vi Í(Vi - V j ) yij* + Vi* ˙ = Pij + jQij 2 ˙˚ ÍÎ j

i

È yij¢* ˘ * S ji = V j Í(V j - Vi ) yij* + V j* ˙ ÍÎ ˙˚ ij

6.8

Computer Techniques in Power System Analysis

Iij

(i)

yij

Vi

Iji

yj

( j) +

y¢ij 2

y¢ij 2

-

+ Sij

Sji

(a)

(

-

(b)

2

Sij + S ji = Vi + V j

2

)

Ê * yij¢* ˆ * * * Á yij + ˜ - ViV j + V jVi yij 2 Ë ¯

(

)

= Pij ,looss + jQij ,loss yij = gij + jbij and

yij 2

= jbii , Pij ,loss = gij Vi - V j

(

2

Qij ,loss = -bii Vi + V j

2

2

)-b

ij

Vi - V j

Vi Vj

2

gij Vi Vj bii

bij

P-V Bus S i = P i + jQ

i

Qgi = Q i + Q + Q + Q Qi i (i)

i

Slack Bus

Sgi

Line 2

i Sgs = S i + S + S + S Si i s S

s

S�i

Power Flow Studies

Y

DVi

(k )

(

= Vi

(k )

- Vi

( k -1)

PQ

)

α

(

a DVi( k ) = a Vi( k ) - Vi( k -1)

)

PQ Vi( k ) = Vi( k -1) + a DVi( k ) α α

α Vi( k )

DVi( k ) = b Re ÈÎ DVi( k ) ˘˚ + j g Im ÎÈ DVi( k ) ˘˚ Vi( k ) = Vi( k -1) + DVi( k ) β

γ Y

RX

Z

Y

Z Z Z

n Fx

0

N

F

x x

n fi x x

xn

i

n

xn

6.9

6.10

Computer Techniques in Power System Analysis

(

)

(

)

-1 x(k ) = x(k -1) - ÈÎ J x(k -1) ˘˚ F x(k -1) ; k = 1, 2,

Jx

Fx È ∂f1 Í ∂x Í 1 Í ∂f 2 Í J ( x ) = Í ∂x1 Í Í Í ∂f n ÍÎ ∂x1

∂f1 ∂x2

∂f1 ˘ ∂xn ˙ ˙ ∂f 2 ˙ ∂xn ˙˙ ˙ ˙ ∂f n ˙ ∂xn ˙˚

∂f 2 ∂x2 ∂f n ∂x2

AD x = b xk

xk

x

= Dx k

(

)

F x ( k -1) = - J ( k -1) D x ( k ) xk =xk

+ Dx k Dx k

Y x

k

xi i n

n PQ n +n

n PV

n n

I i = Â YijVij

i = 1, 2,

,n

J =1

È n ˘ Pi = Re ÍVi  Yij*V j* ˙ i = 1, 2, ÍÎ j =1 ˙˚

,n

È n ˘ Qi = Im ÍVi  Yij*V j* ˙ i = 1, 2, ÍÎ j =1 ˙˚

,n

6.11

Power Flow Studies

Yij Vi = Vi –q i ;

q ij = q i - q j ; Yij = Gij + j Bij

n

Pi - Vi

 ((Gij cos qij + Bij sin qij ) V j ) = 0;

i

n

 ((Gij sin qij - Bij cos qij ) V j ) = 0; i

n

j =1 n

Qi - Vi

j =1

PQ PV

Vi

qi n

PQ

n +n n +n

x È q i each P - Q bus ˘ ( n1 )

x = ÍÍ qi each P - V bus ˙˙ (n2 ) ÍÎ Vi each P - Q bus ˙˚ ( n1 )

x P isp

Vi

sp

n +n Vs sp qs PV

P isp

È Vs Íq Í s Í Pi sp y = Í sp Í Qi Í sp Í Pi Í sp Î Vi

at swing bus ˘ at swing bus ˙ 2 ˙ at P - Q bus ˙ ˙ 2n at P - Q bus ˙ 1 ˙ at P - V bus ˙ ˙ 2n2 at P - V bus ˚

Q isp

y

n +n ÈEquation (6.43) with Pi = Pi sp for each P - Q and P - V bus ˘ ( n1 + n2 ) Í ˙ ÍÎEquation (6.44) with Qi = Qisp for each P - Q bus ˙˚ ( n1 ) n +n Fx

n +n

x

6.12

Computer Techniques in Power System Analysis

(n1 + n2 ) È DP ˘ È0˘ = (n1 ) ÍÎ DQ ˙˚ ÍÎ0˙˚ DP

DQ

Ê n D Pi = Pi sp - Pi = Pi sp - Vi Á Â (Gij cos q ij + Bij sin q ij ) V j Ë j =1

ˆ ˜ ; i = 1, 2, ¯ iπs

,n

Ê n ˆ DQi = Qisp - Qi = Qisp - Vi Á Â (Gij sin q ij - Bij cos q ij ) V j ˜ ; i = 1, 2, , n Ë j =1 ¯ iπs iπP V

È n1 + n2 È DP ˘ =Í H Í DQ ˙ Î ˚ ( k -1) ÍÎ M

n1 ˘ N˙ L ˙˚

( k -1)

È Dq ÍD V Í ÍÎ V

˘ ˙ ˙ ˙˚ ( k )

Èq ˘ Èq ˘ È Dq ˘ +Í ÍV ˙ = ÍV ˙ ˙ Î ˚ ( k ) Î ˚ ( k -1) Î D V ˚ ( k ) H N M ∂Q ∂V

N

PQ n +n

¥n

Vi

for i π j

Hij = Lij Nij

for i = j

Vi

L PQ

H L

∂P ∂P ∂q ∂ V

L

Mij

PV n +n PV n N L n ¥ n +n DV

Vi Vj Gij

qij Bij

Vi Vj Gij Hii

qij + Bij

Qi Bii Vi

Lii = Qi Bii Vi Nii = Pi + Gii Vi

qij qij

∂Q ∂q

Power Flow Studies

6.13

Mii = Pi Gii Vi Ê n ˆ Pi = Vi Á Â (Gij cos q ij + Bij sin q ij ) V j ˜ Ë j =1 ¯ Ê n ˆ Qi = Vi Á Â (Gij sin q ij - Bij cos q ij ) V j ˜ Ë j =1 ¯ Dq

DV V

q

V PV

Qi

Qgi PQ PV

Qgi

Fx

6.14

Computer Techniques in Power System Analysis

q V

P P

Q N

V

q

Q M=0 DP k

Dq k

=Hk

ÈD V ˘ DQ( k -1) = L( k -1) Í ˙ Î V ˚( k ) H

L

Hij = Lij

Vi Vj Gij Hii

qij Bij

qij

iπj

Qi Bii Vi

Lii = Qi Bii Vi

Dq k

ÈD V ˘ ÍÎ V ˙˚ (k )

Dq k q

q q

qk = qk

+ Dq k ÈD V ˘ Í V ˙ Î ˚( k )

L

V

H

L

qij ª Gij

qij tcl

x will settle

Stability Studies

8.7

down to a steady-state value. For different assumed clearing times tcl or sequence of switching operations, the swing curves are computed. From these swing curves it is ascertained whether the system is stable or not. Repetitive computations of the operation sequence.

The equations of the form of Eq. (8.6) representing the dynamics of the synchronous

state variables are: the generator rotor angle (δ), the rotor speed deviation (Dωr), the q E′q e.m.f. (Efd). E q′ represented by a voltage source behind a transient reactance, as shown in Fig. 2.1. The magnitude and the angle of the source voltage are functions of the state variables and the generator terminal bus voltage as shown in Eq. (2.7). Further reduction of this model with certain assumptions gives the classical model as described in Eqs (2.21) – (2.22). In this model, the state variables are the generator rotor angle and the rotor speed deviation. The generator is represented by a voltage of constant magnitude behind the transient reactance and the angle of the voltage is nothing but the rotor angle (δ shown below. The assumptions on which the classical model can be used to represent When the steady-state equilibrium of a power system gets disturbed, there is momentarily a difference between the mechanical power input and the electrical power output in each synchronous machine. This tends to accelerate or decelerate the rotor depending on whether this difference is positive or negative. The mechanical power input gets adjusted according to the governor action. The synchronous ith machine, we can write d (WK.E,i ) + Pdi = Pmi - Pei dt where, WK.E,i = kinetic energy in MW sec; Pdi = damping power in MW

(8.12)

Pmi = mechanical input power in MW; Pei = electrical output power in MW The synchronous frequency is denoted by f0. Since the kinetic energy is proportional to the square of the frequency, we have Ê fi ˆ 0 0 WK.E ,i = WK.E ,i Á Ë f 0 ˜¯

2

where WK.E,i = kinetic energy at synchronous frequency.

(8.13)

8.8

Computer Techniques in Power System Analysis

Since fi = f0 + Dfi, we have WK.E ,i = WK0.E ,i

( f0 + Dfi )2 f 02

Ê 2 Dfi ˆ @ WK0.E ,i Á1 + Ë f 0 ˜¯

since Dfi we have

(8.14) WKiE,i

0

WK.E ,i d d WK.E ,i ) = 2 Dfi ( dt f 0 dt

(8.15)

d d i dq i = - w 0 = w i - w 0 = 2p ( fi - f 0 ) = 2pDfi dt dt where qi is the absolute rotor angle of machine i di is the rotor angle of machine i measured with respect to a synchronously rotating reference frame. Both qi and di are measured in radians. The change in frequency Dfi wi – w0 = 2pDfi. We shall take the damping power to be proportional to the frequency deviation from the synchronous frequency. With these assumptions, substitution of Eq. (8.15) in Eq. (8.12) and division by base MVA yields Let di = qi – w0t and

H i d 2d i dd + Di i = Pmi - Pei , p f 0 dt 2 dt where H i =

0 WK.E ,i

Base MVA

i = 1, 2, … , m

(8.16)

= inertia constant of machine i in p.u.;

Di = damping constant in p.u.; Pmi, Pei are the mechanical and eletrical power in p.u. for machine i; m = number of machines ; The mechanical shaft input power to the generator is a function of the governor in the d–q duration is concerned (of the order of 1–3 s), we can make a number of physically

Fig. 8.3 for just one machine. If there are several machines in the system, each through the transmission network equations. 1. The network is modeled as a lumped circuit in the phasor variables. The implication here is that the nominal AC frequency is much larger compared to the frequencies of the machine oscillations under transient condition. This results in an algebraic description of the network in terms of the phasor variables. Strictly, the network must also be modeled in terms of differential of the normally desired accuracy.

Stability Studies

-

Pe

+

Pm

K 1+Ts

w

1 s

d

8.9

Generator Equations Vf Differential Equations

Governor and Turbine

Exciter

Algebraic Equations Transmission Network

p 0D

1

2. The generator is represented by a voltage source of constant magnitude called the direct-axis transient reactance x¢di model of the synchronous machine. swing. Hence, Vf remains constant. 4. Power output from the governor and the turbine Pm remains essentially 5. The mechanical angle di of each machine rotor coincides with the phase angle of the voltage behind the transient reactance. This again stems from an 6. The loads are converted to an equivalent admittance based on voltages at the The above assumptions simplify the treatment of the transient stability problem. We now have, for the m machine system, the following equations: H i d 2d i dd + Di i = Pmi - Pei , p f 0 dt 2 dt

i = 1, 2, … , m

(8.17)

Pmi Pei is determined from the solution of the algebraic network equations . We now introduce the state variables. xi = di

(8.18)

dqi d d i = + w 0 = w i i = 1, … , m (8.19) dt dt xi represents the rotor angle with respect to a synchronously rotating reference frame, and xi + m, the frequency of the ith machine. xi + m =

8.10

Computer Techniques in Power System Analysis

dd i = wi - w0 dt

(8.20)

dwi D pf = - i p f 0 xi + 0 ( Pmi - Pei ) i = 1, 2, … , m dt Hi Hi If we assume for the time being that Pei di (i = 1, 2, ..., m), then we have 2m to Eqs (8.20) and (8.21) of the form . x = f (x)

(8.21)

(8.22)

More generally, Pei is obtained by solving at each time instant, a set of nonlinear differential algebraic equations of the form x. = F (x, y ) (8.23) 0 = G(x, y)

(8.24)

If we assume loads as constant impedances, then it is possible to eliminate the algebraic variables y and have a mathematical description equivalent to Eq. (8.22) only. This will be discussed in the following sub-section.

Consider a transmission system having n m generator buses (m < n) having a multiport representation as shown in Fig. 8.4. Let Ybus m buses are designated as generator buses. I1

1

+ I2

2

+

Ybus

Im V1

m

+ V2 Vm

-

-

-

In + Vn -

t = 0– will yield the following information:

n 0

(Includes load impedances)

Stability Studies

(a) PLi, QLi, PGi, QGi (b) PLi, QLi, (c) |Vi| –yi = Vi

8.11

i = 1, 2, ..., m i = m + 1, ..., n i = 1, 2, ..., n

P and Q are the real and reactive powers respectively, at the buses and the subscripts L and G refer to the load and generation, respectively. Vi is the phasor voltage at the ith bus. 2. Loads are converted into equivalent admittances yLi between the ith bus and the ground bus as yLi = (PLi – jQLi)/|Vi|2

i = 1, 2, ..., n

(8.25)

If there is no load at a particular bus i, then, yLi = 0. 3. With the addition of elements corresponding to the load admittances, Ybus = Ybus(old) + diag(yLi) n

(8.26)

Ybus. PGi + jQGi is converted into an

the machine as follows Ei = Vi + jx di ¢ Ii = Vi + jx di ¢ ( PGi - jQGi ) = E –d i i Vi*

(8.27)

i = 1, 2, º, m where x di ¢ ith machine and Ei is th the voltage behind the transient reactance of the i machine. The complete representation is shown in Fig. 8.5. jx¢d1

E1 -+

1¢

-+

2¢

+ jx¢d2

I1

I2

jx¢dm Em

m¢

+

2

-

+

E2 -+

1

-

Im -

m

n 0

8.12

Computer Techniques in Power System Analysis

for alternate computation cycles of the differential and algebraic equations, i.e., Eq. (8.23) and Eq. (8.24). It is called the partitioned explicit method (PE). The second method called the simultaneous solution with implicit integration (SI) involves making Eq. (8.23) algebraic by using an implicit integration method and solving these along with Eq. (8.24), simultaneously at each time step as a set of nonlinear algebraic equations. The PE approach is used in the industry in the production grade programs because of its programming simplicity and robust nature of the algorithm. Because of the alternate solution cycles, it is prone to numerical instability. The SI is better when the system contains both slow and fast dynamics. However, if the algebraic variables ( y) are eliminated from Eqs (8.23) and (8.24) and the resulting description is Eq. (8.22), then we can have straightforward numerical integration. This is the case when the loads are considered as constant impedance type.

In this method, the network equations and the differential equations are solved alternately. The network equations are obtained as follows: Let the generator reactances be denoted by the steadystate v–i relations vk = j xd¢ k ik

k = 1, 2, … , m

Ek = specified = Ek –d k

k = 1, 2, … , m

(8.28)

(8.29)

The network representation is Ibus = Ybus Vbus È I1 ˘ È V1 ˘ ˙ Í Í ˙ ˙ Í Í ˙ Í Im ˙ Í Vm ˙ ˙ = [Ybus ] Í Í ˙ Í I m +1 ˙ ÍVm +1 ˙ ˙ Í Í ˙ ˙ Í Í ˙ Î In ˚ Î Vn ˚

i.e.,

(8.30)

Note that the loads have been included in this Ybus as given by Eq. (8.26). When the generator representation is augmented onto the network representation, the following constraints are evident: i k = Ik

k = 1, 2, ..., m

vk = Ek – Vk

k = 1, 2, ..., m

Ik = 0

k = m + 1, ..., n

(8.31)

Stability Studies

8.13

Inverting Eq. (8.28), we get ik = ykvk with yk = – j/x ¢dk. Using the constraint equations Eqs (8.31) and substituting them in Eq. (8.30), we get (Y11 + y1) V1 + Y12V2 +

+ Y1mVm

Y21 + V1 + (Y22 + Y2)V2 + Ym1V1 + Ym2V2 + Ym + 1, m + 1V1 + Yn1V1 +

+

+ Y1nVn = y1E2

+

+ Y2mVm

+ Y2nVn = y2E2

+

+ (Ymm + ym)Vm +

+ YmnVn = ymEm

+

+ Ym + 1, n Vn = 0

+

+ YnnVn = 0

+

(8.32)

These are the network equations, which can be solved for V1, ..., Vn knowing E1, ... Em. The electrical power output of each machine is given by Pei = Re ÈÎ Ei I i* ˘˚

(8.33)

where Ii = (Ei – Vi)yi

i = 1, 2, ...., m

Eq. (8.33) together with the network equations Eqs (8.32), collectively are represented symbolically as 0 = G(x, y) where y is the set of algebraic variables V1, ...,Vm. For numerical integration, the assumed fault clearing time tcl is divided into intervals of Dt such that kDt = tcl where k is an integer. Assume that the voltages behind the transient reactances have been computed as per Eq. (8.27) based 1. For t = 0+ if a symmetrical three-phase to ground fault occurs at the pth bus, set Vp = 0. Then Eq. (8.32) is solved for the voltages Vi. The generator currents Ik (k = 1, 2, ..., m) are computed and hence, the power output Pei of each generator using Eq. (8.33). 2. Equations (8.20) and (8.21) are then integrated for each machine with the initial conditions di (0) = di from Eq. (8.27) and wi (0) = w0, to obtain di and wi at t = Dt. The value of di at t = Dt is denoted by di = (Dt). 3. At t = Dt, the phasor voltages behind the transient reactances, or equivalently the voltage sources Ei (i = 1, 2, ..., m) are given by Ei(Dt) = |Ei| – di(Dt)

(8.34)

Note that by virtue of assumption 2 in Sec. 8.3, the magnitude of this voltage is held constant at its value at t = 0–, but its phase angle is different. With Ei(Dt) given by Eq. (8.34), Eq. (8.32) is again solved for voltages Vi and hence, Ii(i = 1, 2, ..., m) at t = Dt. The power outputs of all the generators at t = Dt are then computed using Eq. (8.33).

8.14

Computer Techniques in Power System Analysis

4. Equations (8.20) and (8.21) are then integrated for each machine using the values di and wi at t = Dt obtained in step 2 as the initial conditions. Thus values of di and wi at t = 2 Dt are obtained for each machine. 5. Steps 3 and 4 are repeated for obtaining di at t = 3Dt and so on until t = tcl, the fault clearing time. 6. At t = tcl clearing of the fault. The network equations Eqs (8.32) are solved with this di and wi are computed for tcl + Dt. 7. The procedure is continued for about 5 s or so. The variation of the rotor angles δi as a function of time constitutes the swing curves. The nature of the swing curves will indicate whether some or all machines are stable or not. clearing time for which all machines indicate stability is called the critical clearing time, denoted by tcr. In other words, if the fault is cleared earlier than this, the system is transiently stable and if cleared later than this, one or more machines will become unstable. Computation of tcr, therefore, involves repetitive simulations.

The trapezoidal rule for integration of a differential equation . x = f(x), x(0) = x0 Dt [ f ( xn ) + f ( xn +1 )] 2 Using this idea, we convert Eqs (8.23) and (8.24) as

is given by

xn +1 = xn +

xn +1 = xn +

Dt [ f ( xn , yn ) + f ( xn +1 , yn +1 )] 2

0 = G(xn+1, yn+1)

n = 0, 1, 2, ...

(8.35) (8.36)

(8.37) (8.38)

In the above equations, the subscripts n and n + 1, refer to the time instants tn and tn + 1, respectively. Equations (8.37) and (8.38) are a set of nonlinear algebraic equations of the form F ( xn +1 , yn +1 ) = 0

(8.39)

G(xn+1, yn+1) = 0

(8.40)

Applying the Newton–Raphson method, we get at (k + 1) iteration (at t = tn + 1) xn(k++11) = xn(k+)1 + Dxn(k+)1

(8.41)

yn(k++11) = yn(k+)1 + Dyn(k+)1

(8.42)

Stability Studies

8.15

where Dxn(k+)1 and Dyn(k+)1 are obtained from solving the linear equations

( (

È- F xn(k+)1 , yn(k+)1 Í Í -G x ( k ) , y ( k ) n +1 n +1 Î

) )

È ∂F ˘ Í ˙ = Í ∂x ˙ Í ∂G ˚ Í Î ∂x

∂F ˘ ˙ È Dxn(k+)1 ˘ ∂y ˙ Í (k ) ˙ ∂G ˙ ÍÎ Dyn +1 ˙˚ ˙ ∂y ˚ ( k )

(8.43)

Computation is carried out n = 1, 2, ..., knowing the values of x, y at n = 0.

In the previous method, the solution of the network equations is required for the faulted and the post-fault systems in order to obtain the electrical power output at each step of integration. If we are not interested in bus voltages as a function of time, these buses can be eliminated, and a multiport representation for both the faulted and the post-fault networks can be obtained at the internal buses. This is possible if all the loads are represented as constant impedances. In this method, we obtain a multiport representation at the internal nodes of the generators by eliminating all the other for Pei = (i = 1, 2, ..., m) is obtained involving only the di’s (i = 1, 2, ..., m). Each time multiport representation also gets changed. The procedure is as follows: 1. Load impedances are absorbed in the Ybus of the transmission network as discussed in Section 8.4.3. Ybus is augmented by the impedances of the synchronous x di¢ . Let the voltages of the new buses be denoted by E1, ..., Em and admittances of the machines be y1, ..., ym. Thus, m new internal buses are created and the overall internal buses are denoted by 1¢, 2¢, ... m¢). Let Ybus (after inclusion of load impedances) be partitioned as m n-m m ÈY1 Y2 ˘ (8.44) Ybus = n - m ÍÎY3 Y4 ˙˚ where Y1 of dimension m ¥ m corresponds to the buses where generators are connected. Y2, Y3, and Y4 are the other sub-matrices. Let the machine admittance y = diag (y1, ..., ym). Then the new augmented Ybus of dimension (n + m) ¥ (n + m) is symbolically represented as m m n-m aug Ybus

0˘ mÈ y -y Í = m - y Y1 + y Y2 ˙ ˙ Í Y3 Y4 ˙˚ n - m ÎÍ 0

(8.45)

8.16

Computer Techniques in Power System Analysis

The bus admittance at the internal nodes is obtained by applying the network reduction

phase to ground fault at bus p is to be simulated, the row and column corresponding to the pth bus is deleted, before doing the network reduction. Let the elements of the Yij(i, j = 1, 2, ..., m). Since m¢= m, we retain the notation m for the sake of convenience. We are now left with a multiport representation shown in Fig. 8.6. I1

1¢

+ I2

Ybus

+

m¢

(Includes load and machine impedances)

Em -

0

Im

E1 E2 -

-

aug

2¢

+

At any bus i, the current into the system is m

I i = Â Yij E j

(8.46)

j =1

The active power into bus i (which is the active power output of generator i) is given by m

Pei = Re ÈÎ Ei I i* ˘˚ = Ei Yii cos q ii + Â Ei E j Yij cos (q ij - d ij ) 2

(8.47)

j =1 j πi

Yij = |Yij| – qij = Gij + jBij and Ei = |Ei| – di

where

The generator dynamics is given by Eq. (8.17) or Eqs (8.20)–(8.21). Since Pei is di s, the above equations can be integrated straightaway, because the initial conditions are known. di

t = 0+

= d i from (8.27) and

dd i dt

t = 0+

= 0,

i = 1, º, m

(8.48)

Pei in Eq. (8.47), |Ei|(i = 1, 2, ..., m) are held constant conductances are neglected, then Gij = 0 and qij reduces to m

Pei = Ei Gii + Â Ei E j Bij sin (d i - d j ) 2

Pei then

j =1 j πi

(8.49)

Stability Studies

8.17

Gij = Yii cos qij and |Yij| = Bij

since

Pei under the faulted and the post-fault stages can be computed by substituting appropriate values of |Yij| and qij be Peif and P eipf respectively. It now remains to integrate the following two systems of equations.

Faulted State H i d 2d i dd + Di i = Pmi - Peif , 0 £ t £ tcl p f 0 dt 2 dt

(8.50)

Post-fault State H i d 2d i dd + Di i = Pmi - Peipf , t ≥ tcl (8.51) 2 p f 0 dt dt The initial values of the state variables, for solving Eq. (8.50) is obtained as shown below. di

t = 0+

= d i from (8.27) and

dd i dt

t = 0+

= 0,

i = 1, 2, … , m

The solution obtained at t = tcl is the initial values of the state variables for R-K method, can be used to integrate these equations to obtain the swing curves. The swing curves can be calculated for different values of tcl and the critical clearing time can be computed as discussed earlier. Conceptually, both the methods, i.e., the PE method and the direct integration method, highlight some interesting aspects of the problem. In the PE method involving network solution, computation proceeds in alternate cycles and the retention of the identity of all the nodes, including load buses, permits the representation of loads. Although we represented the loads by constant impedance, nonlinear representation of the load is possible. Solution of the network equations poses no problem. Monitoring any of the bus voltages as a function of time is possible and this may transient swings, etc. In the direct simulation method, we have to integrate a set of nonlinear differential equations. These can be easily converted into 2m indicated in Eq. (8.20) and Eq. (8.21) which can then be integrated easily. The direct method also leads naturally to a proper understanding of the qualitative methods,

especially true if interest is focused on the critical clearing time, and the bus voltages during the transient period are not required. The synchronous machine shown in Fig. 8.7 is generating 100 q is 1 + j0 and the tie line reactance

8.18

Computer Techniques in Power System Analysis

is 0.08 per unit on a 100 MVA base. The machine transient reactance is 0.20 p.u. and the inertia constant is 4.0 p.u. on a 100 MVA base. Calculate the changes in phase angle and the speed of the generator for a three-phase to ground fault at bus p for with a time increment of 0.02 s. The time for simulation can be taken as 0.2 s. The frequency of supply is 50 Hz. Vp Vq E¢

x¢d = .20

Ppq

Pt Qt

xpq = .08

Qpq p

q

Line current Ipq is given by

Base Load Flow

I pq =

V p - Vq jx pq

V p I *pq = Ppq + jQ pq Substituting the values of Ipq Vp

(V p* - Vq* ) = P - jx pq

pq

+ jQ pq

Ê V p* - (1 + j 0) ˆ Vp Á ˜ = 1.0 + j 0.75 Ë - j 0.08 ¯ since Ppq = Pt, and Qpq = Qt. If we put Vp = eq + jfp (ep + jfp) [(ep – jfp) – (1 + j0)] = 0.06 – j 0.8 Separating the real and imaginary parts, we get f p = 0.08; e 2p + (0.08)2 - e p = 0.06 Solving the quadratic equation and taking only the real value of ep, we get ep = 1.051. Hence, Vp = 1.051 + j 0.08

Stability Studies

8.19

The machine current is given by It =

Pt - jQt V p*

=

1 - j 0.75 = 1.186– - 32.51 = 1 - j 0.638 1.051 - j 0.08

The voltage behind the transient reactance E¢ is given by E ¢ = V p + j xd¢ I t = 1.051 + j 0.08 + j 0.2(1 – j 0.638) = 1.1785 + j 0.28 = 1.21131 – 13.365° The fault at bus p is simulated by setting the voltage at bus p equal Pe. As the fault is on the terminal of the generator, Pe (0) = 0. The differential equations to be solved are given by Eq. (8.20) and Eq. (8.21). Since δ q – w0t dd = w - w0 dt dw p f0 = ( Pm - Pe ) dt H

since damping D ∫ 0. Pm = 1.0 p.u. throughout the calculations since the governor action is not taken into consideration. The initial conditions are d(0) = 13.365° and w (0) = w0 = 2p f0 = 314.1593 rad/s. For t = 0.02 s Step 1 (predictor step) dd dt dw dt

= w (0) - 2p f 0 = 0 0

= 0

p f0 ( Pm - Pe (0)) H

p ¥ 50 (1 - 0) = 39.27 H d w at t = 0.02 s are obtained as =

d ((01).02) = d (0) + Dt

dd dt

= 13.365 + 0 = 13.365°

0

8.20

Computer Techniques in Power System Analysis

w ((10).02) = w (0) + Dt

dw dt

0

= 314.15926 + (0.02) (39.27) = 314.94466 rad/s

Pe((10).02) = 0 . Step 2 d w and Pe, the derivatives at the end of the step are calculated. Then the average of this and the derivative in step 1 are used to compute improved estimates. d d (1) = w ((10).02) - 2p f 0 = 314.95 - 314 dt (0.02) = 0.95

pf d w (1) = 0 Pm - Pe((10).02) dt (0.02) H = 39.27

(

)

Hence the improved estimates for d and w at t = 0.02 are obtained as d ((02.)02) = d (0) +

Dt Ê d d 2 ÁË dt

+ 0

d d (1) ˆ dt (0.02) ˜¯

180 0.02 ˆ = 13.365 + ÊÁ ¥ ˜ (0 + 0.95) = 13.81511∞ Ë p 2 ¯ w ((02.)02) = w (0) +

Dt Ê d w d w (1) ˆ + = 314.94466 2 ÁË dt 0 dt (0.02) ˜¯ d ((02.)02) in order

to obtain the voltage and generator power at t = 0.02. We can improve the estimates by repeating the corrector step to get a better average. t = 0.04 s by incrementing the time by t = 0.02 s. The procedure is continued until t = 0.10 s, at which point the fault clears. After the clearance of fault at t = 0.10 s, the post-fault condition at t = 0.10 must be evaluated, which is then used for t = 0.10 , since the fault is removed, It =

E ¢ - (1 + j 0) j (0.2 + 0.08)

where E ¢ = 1.21131 – 24.615° is obtained from calculation at t = 0.10– s. It is calculated as 1.8019 – j0.36154.

Stability Studies

8.21

Hence, V p = E ¢ - j xd¢ I t = 1.02892 + j 0.14415 Therefore, Pe 0.10+ = Re ÈÎV p I *p ˘˚

(

)

= 1.80188 From this point onwards, the calculations for t = 0.12 s are carried out, by repeating steps 1 and 2 (predictor and corrector steps) using Pe as above at t = 0.10+. Computations are carried out until t = 0.2 s and the values are tabulated as in Table 8.1.

Generator values

Real power Pe p.u. MW

Bus p

Speed rad/s

Voltage magnitude |E¢|p.u. volts

Phase angle d degrees

0

314.1593

1.2113

13.3657

0

0

0

0.0200

314.9447

1.2113

13.8157

0

0

0

0.0400

315.7301

1.2113

15.1657

0

0

0

0.0600

316.5155

1.2113

17.4157

0

0

0

0.0800

317.3009

1.2113

20.5657

0

0

0

0.1000

318.0863

1.2113

24.6157

0

0

0

0.1200

317.3375

1.2113

28.7548

2.811

1.0312

–9.2908

0.1400

316.3954

1.2113

31.9103

2.2867

1.0245

–10.2855

0.1600

315.3213

1.2113

33.8938

2.4124

1.0200

–10.9068

0.1800

314.1795

1.2113

34.5898

2.4558

1.0183

–11.1240

0.2000

313.0355

1.2113

33.9579

2.4164

1.0198

–10.9268

Time t s

Voltage magnitude |Vp| p.u. V

Phase angle dp degrees

(b) Solution by Runge-Kutta Method h = 1. The Runge–Kutta method requires the calculation of four estimates for changes in the phase angles (k1, k2, k3, k4) and in the speed ( 1 , 2 , 3 , 4 ) increment. For t = 0, the initial conditions are known. For t = 0.02, Step 1 First Estimate:

1

=

k1 =

dd Dt = (w (0) - 2p f 0 ) Dt = 0 dt (0)

pf dw p ¥ 50 Dt = 0 ( Pm - Pe (0)) Dt = (1 - 0) ¥ 0.02 = 0.785 dt (0) H 4

8.22

Computer Techniques in Power System Analysis

Step 2 ÊÊ ˆ 1ˆ Second Estimates: k2 = Á Á w (0) + ˜ - 2p f 0 ˜ Dt ËË ¯ 2¯ = ÊÁ ÊÁ 314.16 + ËË 2

=

0.786 ˆ ˆ ˜ - 100p ˜¯ - 0.02 = 0.00785 2 ¯

)

(

p f0 Pm - Pe((10)) Dt H

where Pe((10)) angle d (0) +

k1 . Thus, before 2

2

made. Since the fault is on the terminal of the generator, Pe((10)) = 0 . Therefore, after substitution = 0.785

2

Step 3 Third Estimates: ÊÊ ˆ ˆ k3 = Á Á w (0) + 2 ˜ - 2p f 0 ˜ Dt ËË ¯ 2¯ p f0 Pm - Pe((20)) Dt 3 = H

)

(

where Pe((20)) as in the case of the calculation of second estimates.

d (0) +

k2 . Pe((20)) is zero 2

0.785 ˆ ˆ k3 = ÊÁ ÊÁ 314.16 + ˜ - 100p ˜¯ ¥ 0.02 = 0.00785 ËË 2 ¯ 3

=

p ¥ 50 (1 - 0) ¥ 0.02 = 0.785 4

Step 4 Fourth Estimates Fourth estimates for changes in phase angle and speed are k4 = ((w (0) + 4

=

3

) - 2p f0 ) ¥ 0.02 = 0.01571

(

)

p f0 Pm - Pe((30)) Dt = 0.785 since Pe((30)) = 0 H

Pe((30)) phase angle and speed at t = 0.02 are d (0.02) = d (0) + w (0.02) = w (0) +

d(0) + k3. Finally the 1 (k1 + 2k2 + 2k3 + k4 ) 6 1 ( 6

1

+2

2

+2

3

+

4

).

Stability Studies

8.23

Hence, by substitution, we get d (0.02) = 13.365 +

1 180 = 13.815° (0 + 2 ¥ 0.00785 + 2 ¥ 0.00785 + 0.01571) ¥ 8 p

1 (0.785 + 2 ¥ 0.785 + 2 ¥ 0.785 + 0.785) = 314.945 rad/s 6 d (0.02) in order to compute the voltage and generator power at t = 0.02. This power is zero and will remain zero until the fault is removed. This completes the calculations for t = 0.02. Computations are carried on until t = 0.10, at which point the steps at t = 0.10+ as indicated under the Vp(0.10+) and Pe(0.10+). These are given in the R-K method as w (0.02) = 314.16 +

V p 0.10+ = 1.0289 + j 0.14415

(

)

Pe 0.10+ = 1.80189

(

)

From this point on, the same four steps as discussed earlier, are repeated to obtain the values of d w at t = 0.12 s. Computations are carried until t = 0.2 s. The Runge–Kutta angle and speed for various time intervals from t = 0 to t = 0.2 s are given in Table and the Runge–Kutta method is given in Fig. 8.8. There is very little difference between the two methods.

Bus p

Generator values Time ts

Real power Pe p.u. MW

Speed rad/s

Voltage magnitude |E¢| p.u.

Phase angle d degrees

Voltage magnitude |Vp| p.u.

0

314.1593

1.2113

13.3657

0

0

0

0.0200

314.9447

1.2113

13.8157

0

0

0

0.0400

315.7301

1.2113

15.1657

0

0

0

0.0600

316.5155

1.2113

17.4157

0

0

0

0.0800

317.3009

1.2113

20.5657

0

0

0

0.1000

318.0863

1.2113

24.6157

0

0

0

0.1200

317.3436

1.2113

28.7107

1.9549

1.0313

–9.2769

0.1400

316.4274

1.2113

31.8592

2.1979

1.0246

–10.2695

0.1600

315.3752

1.2113

33.8713

2.3661

1.0200

–10.8997

0.1800

314.2491

1.2113

34.6263

2.4545

1.0182

–11.1353

0.2000

313.1126

1.2113

34.0758

2.4613

1.0195

–10.9636

Phase angle dp degrees

8.24

Computer Techniques in Power System Analysis

318

40

317

35 ME

316

30 RK

? 25

RK

d ME

20

315 314

15

313

10

312

5

311

0

0.04

0.08

0.12

0.16

0.20

Time, t (seconds)

A sample system is described in Fig. 8.9 for which the machine data and the transmission line data are given in per unit in Tables 8.3 and 8.4,

compute the changes in phase angles of generators for a three-phase fault at bus 2 which clears itself (self-clearing fault) at 0.1 s. Take a time step of 0.02 s and the duration of time study as 1.0 s. Take 1 as the slack bus. (b) Repeat the study by the are on 100 MVA base. 1 G1

3

G2 2

Stability Studies

Bus code p 1 2

Inertia constant H in sec. 160 3

Line no. 1 2 3

Bus code p-q 1-2 1-3 2-3

Bus code p

Bus voltages

1 2 3

1.04 – 0° 1.02 – – 3.09° 0.93 – – 7.01°

8.25

Direct axis transient reactance xd¢ in p.u. 0.1 0.3

Half line charging y ¢pq /2 j 0.15 j 0.07 j 0.04

Admittance Ypq 1.47 – j 5.88 2.94 – j 11.77 2.75 – j 9.17

Generation MW MVARs 212.14 93.26 100.0 70.0 0.0 0.0

MW 0.0 50.0 250.0

Load MVARs 0.0 20.0 150.0

1. Conversion of loads into equivalent admittances—Using Eq. (8.25), the loads yL2 = 0.48 – j 0.19; yL3 = 2.91 – j1.75 2. Conversion of scheduled generation to equivalent voltage sources—Using Eq. (8.27), we obtain the voltages behind the transient reactances for machines 1 and 2 as E1 = 1.15–10.24∞;

y1 = - j xd¢ 1 = - j10.0

E2 = 1.26–10.39∞;

y2 = - j xd¢ 2 = - j 3.33

Hence, d1 = 10.24∞ = 0.1786 radians; d 2 = 10.39∞ = 0.181 radianst 3. Partitioned explicit method — By augmenting the generator representation to the network equations corresponding to Eq. (8.32) are obtained as È 4.41 - j 27.43 -1.47 + j 5.88 -2.94 + j11.77 ˘ ÈV1 ˘ È - j10 E1 ˘ Í -1.47 + j 5.88 4.70 - j18.39 -2.75 + j 9.17 ˙ ÍV ˙ = Í- j 3.3E ˙ 2˙ Í ˙ Í 2˙ Í Í ˙ ÍÎ-2.94 + j11.77 -2.75 + j 9.17 8.61 - j 22.58 ˙˚ ÎV3 ˚ ÍÎ 0 ˙˚ At t = 0, the fault at bus 2 is simulated in the above equations by setting V2 ∫ 0 and iteratively solving for V1 and V2 with E1 and E2 2 above. The bus voltages thus obtained are V1 = 0.53 – 0.56°, V2 = 0, V3 = 0.27 – – 6.27°. The generator currents and power outputs of the generators are obtained from Eq. (8.33) as I1 = 1.99 – j6.0; I2 = 0.76 – j 4.13 Pel = 1.0212 and Pe2 = 0.0

8.26

Computer Techniques in Power System Analysis

The differential equations are dd i = wi - w0 dt dw i p f0 = ( Pmi - Pei ) i = 1, 2. dt Hi with the initial conditions d1(0) = 0.1786 rad, d2(0) = 0.181 rad., and w1(0) = w2 (0) = 314.16 rad/s. as follows: d1((10) .02) = 0.1786 rad; d 2(1()0.02) = 0.1814 rad; w1(1(0) .02) = 314.1816; w 2(1()0.02) = 315.2072.

The voltages behind the transient reactances with these di’s are given as per Eq. (8.34) and the network equations solved again for the voltages Vi and hence the powers Pci (i d’s and w Euler’s method are obtained as d1(0.02) = 0.1789 rad = 10.25°; d2(0.02) = 0.1918 rad = 10.99°; w1(0.02) = 314.1816;

w2(0.02) = 315.2072

The voltages behind the transient reactances are now given by E1 = 1.15 – 10.25°;

E2 = 1.26 + – 10.99°

With these voltages, the network equations are again solved to get V1, V2, Pe1, and Pe2 at t = 0.02 s as V1 = 0.53–0.57∞; V2 = 0; V3 = 0.27– 6.26∞; Pe1 = 1.02; Pe 2 = 0. This completes the calculation for t + Dt = 0.02 s. Calculation is then done in a t= 0.10 s, however, the network equations are solved without the fault to obtain the postfault conditions before proceeding with the calculations. Only the magnitudes of the voltages behind transient reactances are held constant. The process is continued until t = 1.0 s. The rotor angles of the generators 1 and 2 are plotted in Fig. 8.10. The system is stable for this disturbance. 4. (Table 8.6) Initial values: E1 = 1.1479 – 10.2354°; E2 = 1.2607 – 10.4016°; The magnitude part remains constant, although in the simulation, the angle, i.e., delta varies. On application of fault: Bus voltage values, V1 = 0.5291 – 0.5551°; V2 = 0; V3 = 0.2655 – – 6.2768°. Real power generated, Pe1 = 1.0212; Pe2 = 0. d1 = 0.1786; d2 = 0.18115; w1 = 314.1809; w2 = 315.2069. (0.02 s),

Stability Studies

d1 = 0.1789 = 10.2478°; d2 = 0.1920 = 11.0018°;

8.27

w1 = 314.1809; w2 = 315.2069;

w2 = 315.2069. Bus voltages at this point, V1 = 0.5291 – 0.5675°; V2 = 0; V3 = 0.2655 – – 6.2645°. Real power generated, Pel = 1.0212; Pe2 = 0. 50

40 2

30 1

20

10

0

0.16

0.32

0.48

0.64

0.8

0.96

Time (sec)

-10 -20 Prefault

Postfault Faulted

This completes the calculation for t = 0.02 s. Proceeding like this, we get the simulations for 1 s (0.1 s faulted and 0.9 s post-fault). In Table 8.6, the values of deltas, omegas, and powers are given in tabular form. Note that the time-steps for both the faulted and the post-fault condition are 0.02 s. However, results are given here at intervals of 0.1 s for the post-fault system to save space.

Generator 1

Generator 2

Time (in sec)

Delta (degree)

Speed (radian)

Power (in p.u.)

0

10.2354

314.1593

1.1215

0.0200

10.2478

314.1809

0.0400

10.2849

0.0600

10.3468

0.0800

10.4334

Delta (degree)

Speed (radian)

Power (in p.u.)

10.4016

314.1593

1.0004

1.0212

11.0018

315.2069

0

314.2025

1.0212

12.8026

316.2545

0

314.2241

1.0212

15.8038

317.3021

0

314.2457

1.0212

20.0055

318.3498

0

Contd.

8.28

Computer Techniques in Power System Analysis

Generator 1

Generator 2

Time (in sec)

Delta (degree)

Speed (radian)

Power (in p.u.)

Delta (degree)

0.1000

10.5449

314.2673

1.0212

25.4077

319.3974

Speed (radian)

Power (in p.u.) 0

0.2000

11.4106

314.3630

0.9991

42.1341

314.2200

2.1465

0.3000

12.8742

314.4564

1.3533

26.3902

309.1705

1.7702

0.4000

14.6102

314.4432

2.6839

– 4.0021

309.8184

0.4633

0.5000

15.9377

314.3305

3.3906

– 13.9766

315.3984

– 0.1709

0.6000

16.6150

314.2400

2.7621

8.2276

319.9207

0.3905

0.7000

17.1014

314.2714

1.4083

39.9316

318.2384

1.7134

0.8000

18.0401

314.3784

0.9829

47.2392

312.4190

2.1644

0.9000

19.5496

314.4504

1.6807

23.5914

308.5492

1.4365

1.0000

21.1361

314.3989

3.0328

– 4.3762

311.1501

0.1427

6. Direct simulation method Ybus, which includes the loads represented as shunt impedances to ground, is augmented by the impedances of the machines. The resulting Ybus is 1¢

2¢

1

2

3

1¢ È- j10.0 0 j10.0 0 0 ˘ Í ˙ 2¢ - j 3.33 + j 3.33 0 0 0 Í ˙ 1 Í+ j10.0 0 4.41 - j 27.43 -1.47 + j 5.88 -2.94 + j11.77 ˙ Í ˙ 2 Í 0 + j 3.33 -1.47 + j 5.88 4.7 - j18.39 -2.75 + j 9.17 ˙ 3 ÍÎ 0 0 -2.94 + j11.77 -2.75 + j 9.17 8.61 - 22.58 ˙˚ Since the fault is at bus 2 with zero impedance to the ground, it is simulated by deleting the row and column corresponding to bus 2 in the resulting Ybus. The bus 1¢ and 2¢ during faulted condition is obtained by applying the network reduction formula discussed in Chapter 3. 1¢ 2¢ 1¢ È0.78 - j 5.46 0 ˘ 2 ¢ ÍÎ 0 - j 3.33˙˚ The swing equations during the faulted state are d d1 dd 2 = w1 - 314.16 ; = w 2 - 314.16 dt dt d w1 p ¥ 50 = (2.1214 - 1.0212) = 1.08 ; 160 dt

d w 2 p ¥ 50 = (1 - 0) = 52.36 3 dt

The initial conditions at t = 0+ are d1 = 10.24° = 0.1786 rad; d2 = 10.39° = 0.18 rad;

Stability Studies

8.29

w1 = w2 = 314.16 rad/s = w2 integrated until t = 0.1 s. The values of d1 and wi are shown in Table 8.8 for i = 1, 2. For the post-fault state, there is no fault at bus and hence, the bus admittance 1¢ 1¢ 2¢

2¢

È1.21 - j 2.92 0.37 + j1.67 ˘ Í0.37 + j1.67 0.29 - j 2.23˙ Î ˚

The post-fault state system swing equations valid for t ≥ 0.10+ are

(

(

))

(

(

))

0 ˘˚ d1 = 0.98 ÈÎ2.12 - 1.6 + 2.48 cos 77.67∞ - d12 0 ˘˚ d 2 = 52.36 ÈÎ1.0 - 0.46 + 2.48 cos 77.67∞ - d 21

faulted state. They are then integrated using the values of di, wi(i = 1,2) at t = 0.1 s. Thus, we integrate the two systems of equations (faulted and post-fault states) by any of the well known numerical methods. The internal voltage angles of generators 1 and 2 are the same as in Fig. 8.8, which were obtained by the PE method. Table 8.8 system is seen to be stable for this disturbance.

Generator 1

Generator 2

Time (in sec)

Delta (degree)

Speed (radian)

Power (in p.u.)

Delta (degree)

Speed (radian)

Power (in p.u.)

0

10.2354

314.1593

2.1214

10.4016

314.1593

1.0000

0.0200

10.2478

314.1809

1.0204

11.0016

315.2065

0

0.0400

10.2849

314.2025

1.0204

12.8016

316.2537

0

0.0600

10.3469

314.2241

1.0204

15.8016

317.3009

0

0.0800

10.4336

314.2457

1.0204

20.0016

318.3481

0

0.1000

10.5450

314.2674

1.0204

25.4016

319.3953

0

0.2000

11.4371

314.3750

0.8295

42.2614

314.2632

2.1396

0.3000

13.0116

314.4804

1.2830

26.8272

309.2282

1.7692

0.4000

14.8937

314.4709

2.6333

– 3.3578

309.8677

0.4543

0.5000

16.4100

314.3703

3.2421

– 11.9772

315.8728

– 0.2648

0.6000

17.3646

314.2974

2.5767

12.8823

320.1428

0.5163

0.7000

18.2307

314.3468

1.2253

43.5595

317.7422

1.8183

0.8000

19.6565

314.4716

0.8318

47.5035

311.8604

2.1379

0.9000

21.6992

314.5388

1.8137

22.0986

308.5942

1.2905

1.0000

23.7597

314.4797

3.0427

– 3.2065

312.0400

–0.0172

8.30

Computer Techniques in Power System Analysis

The FACTS devices described in Chapter 7 are implemented with some suitable controller. The controller modulated the value of some variable/parameter of the system. So, for dynamic simulation, the equations corresponding to the controllers are to be solved to obtain the value of the controlled variable which is then used in the solution of the differential and algebraic equations representing the system. The control structure is chosen depending on the objective of installation of the FACTS device. Hence, for the same FACTS device, the controller equations may be different. Here, the cases of two FACTS devices, one series (TCSC) and one shunt type (STATCOM) are discussed.

The structure of a TCSC was discussed in Sec. 2.7.1, and shown pictorially in Fig. 2.24. TCSC is represented as a variable capacitor, whose capacitance value a of the thyristors. Therefore, the controller used for typical control structure used for the TCSC is shown in Fig. 8.11. Here, the active P) through the line (where the TCSC is placed) is taken as the control (Pref), P and Pref are compared and the error is fed to a proportional-integral (P-I) a) of the TCSC. A limiter is placed on the value of a and the output of the limiter is used to calculate the net capacitance of the TCSC, Ctcsc. The difference between the Pref and P is denoted by error. The error calculated in each time step is integrated to get INTerror. The value of INTerror at time t1 is the integration of the errors of all the previous time steps. This is shown in Eq. (8.52).

error (t1 ) = Pref - P; INTerror (t1 ) = INTerror (t0 ) + error (t1 ) ;

(8.52)

a (t1 ) = K P . error (t1 ) + K I .INTerror (t1 )

(8.53)

Generally, the value of a is modulated around some reference value of α, say aref and

a (t1 ) = a ref + K P . error (t1 ) + K I .INTerror (t1 )

Pref

+ P

-

KI KP + s Controller

a

Ctcsc TCSC

(8.54)

Power System Network

P

Stability Studies

8.31

Equation (8.52) and (8.54) are solved in every time step and the value of α is used to get the value of Ctcsc. Ybus as shown in Eq. (8.55). Ybus (i, j ) = Ybus (i, j ) + 1 ( 2p f Ct csc )

(8.55)

Ybus and thus the effect of the TCSC gets included. The control structure may be different depending on the system requirement. However, method of simulation always involves solution

The WSCC 3-machine, 9-bus system is shown in Fig. 8.12

phase fault takes place at bus 6 which clears itself (self-clearing fault) after 0.50 s, compute the value of the generator relative rotor angles (taking gen 1 as reference) at the point of fault clearance. Also compute and plot the variation of the generator relative rotor angles for 2 seconds after the fault clearance. Consider a simulation time step of 1/600 s. Now consider a TCSC controller in the line between buses 5

the generator relative rotor angles (taking gen 1 as reference) at the point of fault clearance and also compute and plot the variation of the generator relative rotor angles for 2 seconds after the fault clearance. obtained as shown below: The current injected by the generator connected to the ith bus, I di = ( Pi + j Qi / Vi – Qi )2

(8.56)

where Pi + jQi is the power injection by the generator and Vi – θi is the bus voltage and its angle. The generator rotor angle, δi is the angle of the voltage

The d from the relation

E i = Vi – Qi + jxqi I i

(8.57)

I di + jI qi = I i e(p

(8.58)

q 2 -d i )

E ¢q) can be

So the q obtained as Eq¢i = Vi cos (d i - q i ) + xd¢ i I di

(8.59)

8.32

Computer Techniques in Power System Analysis

(

)

E fdi = Eq¢i + xdi - xd¢ i I di Vrefi =

E fdi K Ai

(8.60)

+ Vi

(8.61)

The pre-fault values of the state variables of the three machines are listed in Table 8.8.

d (deg)

ws – wr (rad/s)

E¢q (p.u.)

Efd (p.u.)

1

3.5862

0

1.0562

1.0819

2

61.0661

0

0.7886

1.7893

3

54.0844

0

0.7687

1.4034

Machine no

to ground. yL 5 = 1.2605 - j 0.5042;

yL 6 = 0.8773 - j 0.2924; yL8 = 0.9684 - j 0.3389;

The direct method is used to carry out the simulation. The values of the relative rotor angles (δ2 – δ1) and (δ3 – δ1) just at the instant of fault clearance are enlisted in Table 8.10. The plot of the relative rotor angle (δ2 – δ1) for 2 seconds after the fault as described in Sec. 8.6.1. The controller constants selected are KP = 0.8;

d2 - d3 (deg)

1.5

KI = 0.12;

Without TCSC With TCSC

1.0

0.5

0.0 4.5

5.0

5.5

6

Time (s)

Simulation is again carried out and the values of the relative rotor angles (δ2 – δ1) and (δ3 – δ1) just at the instant of fault clearance are enlisted in Table 8.9. The plot of the relative rotor angle (δ2 – δ1) for 2 seconds after the fault clearance is shown in Fig. 8.12 for comparison.

Stability Studies

Relative rotor angle at fault clearance

8.33

System without TCSC System with TCSC

(δ2 – δ1)

57.4800

55.1850

(δ3 – δ1)

50.4982

49.0341

The structure of the STATCOM was shown in Sec. 2.7.2, (Fig. 2.55). It consists of a voltage sourced converter (VSC), connected to a system bus through some coupling transformer. Usually, the VSC is represented by a controllable voltage source. The STATCOM is mainly used for the reactive power support and to maintain the bus voltage. Therefore the control variable might be either the reactive power injection or the voltage of the bus where the STATCOM is connected. Various control strategies can be used to control the output voltage of the VSC and thus to control the reactive power supplied by it. For simulation, just like the case of TCSC, the controller equations are solved in each time step to obtain the VSC voltage (magnitude and angle). This voltage is then included in the solution of the network.

of multi-machine systems when subjected to large and sudden disturbances. Since

(2) the direct integration method. Method (1) has the advantage that voltage variables at various buses could be monitored during the disturbance. In method (2) that is quicker, this information is missing. The numerical schemes can be either the used in transient stability analysis using Energy Function method.

8.1

H= 3.0 s, x ¢d = 0.25 p.u., line reactance x = 0.10 p.u. Initially the generator bus voltage is 1.40 p.u. and delivers a power of 75 MW. At t = 0, there is a threephase short-circuit at the generator bus which is cleared in 0.1 s. The base – 0°. (a) Write the swing equations for the faulted and the post-fault states with all D = 0.1 p.u.

fourth-order R-K method. 8.2 A synchronous machine having an inertia constant of H = 5.00 s and transient reactance of 0.25 p.u. delivers power of 25 MW over two parallel transmission

8.34

Computer Techniques in Power System Analysis

lines each having reactance of 0.20 p.u. The other end of the transmission line – 0°. The voltage behind the transient reactance of the generator is 1.025 p.u. A three-phase short-circuit occurs at the middle of one of the lines and is cleared in 0.3 s by simultaneous opening of the circuit breakers at both ends of the transmission line. From the swing equations for the faulted and the post-fault state, compute method. Use 50 as the base MVA. 8.3 Repeat Problem 8.2 if the fault is at the generator end of the line. 8.4 P.8.1).

3

1

4

5

2

The line data are as follows: Line data bus code

Impedance p.u.

1-2

0.02 + j 0.06

1-3

0.08 + j 0.24

j 0.025

2-3 2-4 2-5 3-4 4-5

0.06 + j 0.18 0.06 + j 0.18 0.04 + j 0.12 0.01 + j 0.03 0.08 + j 0.24

j 0.02 j 0.02 j 0.015 j 0.01 j 0.025

Bus number

Half-line charging admittance p.u. j 0.03

Generation

Bus voltage

MW

Load

MVAR

MW

MVAR

1

1.06 + j 0.0

129.565

– 7.48

0

0

2

1.04621 – j 0.05128

40.0

30.0

20

10

3

1. 02032 – j 0.0892

0

0

45

15

4

1.01917 – j 0.09506

0

0

40

5

5

1.01209 – j 0.10906

0

0

60

10

Stability Studies

8.35

Choose 100 as the base MVA. The synchronous machine data on the common 100 MVA base is as follows: Machine connected to bus 1 H = 50 s xd¢ = 0.25 p.u. Machine connected to bus 2 H = 7.0 s xd¢ = 1.50 p.u. The fault is at bus 2 for a duration of 0.1 s. (a) Set up the equations for dynamic simulation by the alternate cycle solution method, i.e., Eqs (8.15), (8.16), (8.36), and (8.37). (b) Choosing a time increment of t d1, d2, w1, w2, and the bus voltages at t + t integration method. (c) Obtain a two-port representation at the internal nodes of the generators connected to buses 1 and 2, and by the direct method, obtain d1, d2, w1, and w2 at t the fourth-order R–K method. 8.5 Write a computer program for the transient stability evaluation of a singlefor the faulted and the post-fault state to be given. Use this program to verify the results of Problems 8.1, 8.2, and 8.3. 8.6 For the transient stability of a multi-machine system, write a computer program using both (a) alternate cycle solution method, and (b) direct method. The network reduction needed in the program must form one of the subroutines. program. 8.7 Write a computer program to integrate a system of differential equations x = f (x using the program, integrate the following differential equations:

(

)

(a). x + 4 x 2 - 1 x + x = 0; (b). x + 2 xx + x = 0;

x (0) = 0.75, x (0) = 0

x (0) = 1, x (0) = 0

Use time increment t = 0.05 s for (a) and 0.1 s for (b).

9 Utilization of the mechanical power available in wind through wind mills to carry out different jobs is not new. Subsequently, small generators were connected to the wind turbines to convert the mechanical power to electrical form. From the initial stages till thirty years back, these generating facilities were isolated and catering to small local loads. Afterwards, large numbers of such wind turbine-generator combinations were installed in the same location, thus forming a wind farm. As the search for candidate as the technology was reasonably mature and the cost of generation was very much affordable. So, people started looking for large sites with available wind speed beyond the threshold (4 m/s) and new and larges wind farms were established. The generation of these farms started exceeding 100 MW and is approaching GW levels. Naturally, local loads are not big enough to consume this power and the question of connecting to the grid to transmit the power to other locations has become necessary. Wind speed being variable in nature, the integration of this power with the grid poses a technical challenge. Hence, modeling of the wind turbine and generators for including them in power system studies becomes important. concept), where wind energy is transformed into electrical energy using a simple squirrel-cage induction generator (SCIG) directly connected to the power grid as gearbox. At all the operating points, this turbine operates at a constant speed. Usually reactive power drawn by the singly-excited induction generator (SEIG). However, the energy available in the wind. In this type, turbine speed is adjusted as a function of the wind speed to maximize the output power. A synchronous generator can be used with this turbine. However, this produces three-phase power of variable voltage and frequency. Therefore, back to back converters are required to change the voltage and the frequency level to match the grid frequency (Fig. 9.1(b)). The converters have to handle the rated power of the plant, which makes it costly. needed for the rotor circuit handling comparatively less power (typically 30%). The scheme is shown in Fig. 9.1(c). The reactive power compensation (through shunt connected capacitor) is not required in this case.

9.2

Computer Techniques in Power System Analysis

G R I D

SCIG Wind Turbine

Gear Box

Transformer

G R I D

Synch Gen Wind Turbine

Gear Box

Machine side Converter

Grid side Transformer Converter

G R I D

DFIG Wind Turbine

Gear Box

Transformer

Rotor side Converter (RSC)

Grid side Converter (GSC)

also has few disadvantages like: (a) the presence of slip ring and gear box gives rise to maintenance issues, (b) individual machine usually not available with a rating higher than 5 MW, thus making it necessary to use larger number of machines in a wind farm. These problems can be overcome with the use of permanent magnet synchronous machine (PMSM) though these are very costly at this point of time.

Similar to thermal or hydro generation, wind power generation also has two broad components: (a) the electrical part, i.e. the generator, and (b) the mechanical part which includes mainly the turbine and the generator rotor. The models of these

Renewable Sources—Wind Energy System

9.3

components suitable for inclusion in power system studies are discussed here. But before that, the notations used for different system variables and parameters while describing the models are given below: and qs = the stator d-axis and q-axis voltages respectively and qr = the rotor d-axis and q-axis voltages respectively ids and iqs = the stator d-axis and q-axis currents respectively iqr and idr = the rotor d-axis and q-axis currents respectively e¢d and e¢q = the equivalent d-axis and q-axis voltages behind transient impedance respectively Tm and Te = the mechanical torque and the electromagnetic torque respectively lm = the mutual inductance, lss and lrr are the stator and the rotor inductances rs and rr = the stator and the rotor resistance respectively eB = the electrical base speed r and t = the rotor electrical and mechanical speed respectively tw = the shaft torsional angle (twist angle) Ht and Hg = the turbine and the generator inertia ksh = the drive train shaft stiffness Csh Cp = tip speed ratio of the wind turbine = pitch angle of the wind turbine R = the blade length of the wind turbine = the air density Vw = the wind velocity The d-axis and q-axis mentioned above are those of the synchronously rotating reference frame. The convention considered here is that the q-axis leads the d-axis. ds dr

The wind turbine, the wind generator rotor shaft, and the intermediate gear box can be represented by a 5th order model. However, generally a comparatively reduced order model termed as two-mass model two-mass model are given as dw r 1 = [kshqtw + Cshw eB (w t - w r ) - Te ] dt 2H g

(9.1)

dq tw = w eB (w t - w r ) dt

(9.2)

dwt 1 = [Tm - kshqtw - Cshw eB (w t - w r )] 2Ht dt

(9.3)

9.4

Computer Techniques in Power System Analysis

The mechanical torque Tm is given by Tm =

0.5rp R 2 C p (l , b )Vw3 Ê 1 ÁË T w t

ˆ ˜ mBase ¯

(9.4)

lumped mass or single-mass model may be used. Here, the mechanical system is represented by only one differential equation as shown below. Since the turbine, the gear box, and the generator rotor is considered as a single-mass, ωr = ωt and the equivalent inertia of the lumped mass, Heq = Hg + Ht. dw r 1 = [Tm - Te ] dt 2 H eq

(9.5)

To avoid damages due to very high wind speeds, a provision of controlling the pitch angle (β) of the wind turbine blades is there. Till the generator speed is within the rated value (wrrated), the pitch angle is maintained at 0 and the torque of the machine is controlled such that maximum available power at the given condition (i.e., wind speed) is extracted. The torque for this maximum power tracking can be obtained using the relation

where Cp

max

Cp = Cp

Te = K opt w r2

(9.6)

3 K opt = 0.5rp R5C p max w t3rated lopt Prated

(9.7)

is the maximum value of Cp when β = 0 , λopt is the tip speed ratio when .

max

Cp and λopt are constants of the wind turbine and can be obtained from the max turbine data. When the generator speed becomes greater than or equal to the rated speed wr , rated the pitch angle is controlled and the torque is maintained constant at the rated value, Te . rated

As mentioned in the previous section, the most popularly used machine for the conversion of wind energy to electrical form is the induction machine (IM). Considering that both the stator and the rotor windings of the IM are excited (i.e. doubly-fed induction generator), the stator and the rotor circuits can be represented by the following equations: dids w eB = dt ls¢

eq¢ È Ê ˘ (lss - ls¢ ) ˆ Í- ÁË rs + T ˜¯ ids + ls¢iqs - T + w r ed¢ + K mrr vdr - vds ˙ Î ˚ r r

(9.8)

diqs

ed¢ (lss - ls¢ ) ˆ È Ê ˘ Í- ÁË rs + T ˜¯ iqs + ls¢ids - T + w r eq¢ + K mrr vqr - vqs ˙ Î ˚ r r

(9.9)

dt

=

w eB ls¢

ded¢ e¢ È (l - l ¢ ) ˘ = w eB Íw s ss s iqs - d + w s (w s - w r ) eq¢ - w s K mrr vqr ˙ dt Tr Tr Î ˚

(9.10)

Renewable Sources—Wind Energy System

eq¢ È ˘ (l - l ¢ ) = w eB Í-w s ss s ids - - w s (w s - w r ) ed¢ - w s K mrr vdr ˙ dt Tr Tr Î ˚

deq¢

(

where l ¢ = lss - lm2 lrr

)

9.5

(9.11) (9.12)

Kmrr = lm/lrr;

(9.13)

Tr = lrr/rr

(9.14)

ed¢ = w s lm (iqr + K mrr iqs )

(9.15)

eq¢ = -w s lm (idr + K mrr ids )

(9.16)

Here, e¢d and e¢q are actually voltages proportional to the d-axis and the q rotor side converter (RSC), as shown in Fig. 9.1(c). So vdr and vqr are the d-axis and the q-axis components of the output voltage of the RSC. Generally, the dynamics of the stator and the rotor windings, given by Eqs (9.8) – (9.11), are much faster than the dynamics of the mechanical system and hence, for power system application, usually the stator and rotor dynamics are neglected. This means setting the differential terms in Eqs (9.8) – (9.11) to zero and converting them to algebraic equations as given below. vds = - rs ids + w s ls¢iqs + ed¢ = - rs ids + w s (lss iqs + lm iqr )

(9.17)

vqs = - rs iqs - w s ls¢ids + eq¢ = - rs iqs - w s (lss ids + lm idr )

(9.18)

vdr = - rr idr + (w s - w r ) (lrr iqr + lm iqs )

(9.19)

vqr = - rr iqr - (w s - w r ) (lrr idr + lm ids )

(9.20)

Ptot = Ps + Pr = vds ids + vqs iqs + vdr idr + vqr iqr

(9.21)

Qtot = Qs + QGSC = vqs ids - vds iqs

(9.22)

expressed as

The grid side converter (GSC) is operated at unity power factor and hence, QGSC = 0. The electromagnetic torque Te is given by Te = (eq¢ iqs + ed¢ ids ) w s = Lm (ids iqr - iqs idr )

(9.23)

The wind power generation usually takes place at a voltage much less than the transmission system voltages. Also, the site is usually distant from the existing system buses. Hence, the wind generator is connected to the power system buses through a

9.6

Computer Techniques in Power System Analysis

transformer and a distribution line. Sometimes, local loads are also connected at the wind generator bus. A typical connection arrangement is shown in Fig. 9.2.

the presence of the back to back converters in the rotor circuit (Fig. 9.1(c)). So in a P and Q Fig. 9.2, then the values of Ptot and Qtot are obtained by subtracting the losses in the connecting line and the transformer impedance (r + jx) from P and Q. Thus, solving

Ptot + jQtot DFIG

Pt

r + jx

Ps, Qs

To System

1 : ns Pr, Qr

PGSC, QGSC RSC

GSC

Original load PL, QL Local load PDG, QDG

reference, the values of vds and vqs are the real and the reactive part of the bus voltage respectively. So, in the six equations, Eqs (9.17) – (9.22), there are 7 unknowns, namely, (ids, iqs, idr, iqr, vdr, v and ωr). Hence, one more equation is needed for the solution, which can be obtained using the two expressions of torque given in Eq. (9.6) and Eq. (9.23) as shown below. Lm (ids iqr - iqs idr ) = K opt w r2

(9.24)

These seven equations, Eqs (9.17) – (9.22), and Eq. (9.24) are solved using the Newton-Raphson method to get the stator and rotor currents, rotor voltages and the rotor speed. These are used to obtain the value of Te. Under steady-state, Tm = Te ; ωt = ωr ; θtw = Tm / ksh.

bus 5 of the power system described in Example 6.2 through a transformer and a distribution line as shown in Fig. 9.3. The combined impedance of the transformer and the distribution line is (r + jx) = (0.003 + j0.02) p.u. At a particular time, the wind farm injects 25 MW active power (and no reactive power) at bus 5. An additional

Renewable Sources—Wind Energy System

9.7

rating 5 MW is given in Appendix C. The data is on the own base of the machine. in parallel. Since the system base MVA is also taken as 50 MVA (as in Example 6.2), machine data. 1

4

3

a:1

Gen 1 Load a:1 6

5 Original Load

Local Load

Load

2

Gen 2

DFIG

admittance matrix is formed. YBUS = È0.9922 0 Í - j 4.3925 Í 1.0214 Í 0 Í - j1.9545 Í 0.4449 Í 0 Í + j 0.66461 Í 0 5583 . Í 0 Í + j 2.5820 Í - 0.5765 Í 0 Í + j1.3085 Í Í - 0.4339 0 Í + j1.8275 Í Í 0 0 Í Î

0 - 0.4449 + j 0.6461 0.4449 - j 8.1649 0 + j 6.8353 0 0 0

- 0.5583 + j 2.5820 0 0 + j 6.8353 1.1124 - j11.1062 0 - 0.5541 + j 2.3249 0

0 - 0.5765 + j1.3085 0 0 7.9115 - j 53.5416 0 + j 3.2520 -7.3350 + j 48.8998

- 0.4339 + j1.8275 0 0 - 0.5541 + j 2.3249 0 + j 3.2520 0.9880 - j 7.3075 0

The power injected at bus 5 = 25 MW + 0 MVAR = (0.5 + j0.0) p.u.

˘ ˙ ˙ ˙ 0 ˙ ˙ ˙ 0 ˙ ˙ ˙ 0 ˙ ˙ -7.3350 ˙ + j 48.8998˙ ˙ ˙ 0 ˙ ˙ 7.3350 ˙ - j 48.8998˙˚ 0

9.8

Computer Techniques in Power System Analysis

From the converged solution of the system without wind farm (Table 6.7), the voltage of bus 5, |V5|–q5 = 0.8511 – – 13.03° Idg| – fgd = (0.5/0.8511 – – 13.03°)* = 0.5875 – – 13.03° Voltage at bus 7, |V7| – q7 = 0.8511 – – 13.03° + j(0.003 + j 0.02) (0.5875 – – 13.03°) = 0.8521 – – 12.04° Hence, vds = 0.832; vqs = – 0.184; Power at bus 7 = Ptot + jQtot = |V7| – q7 (|Idg| – fdg)* = (0.8521 – – 12.04°) (0.5875 – – 13.03°) x = 0.501 + j 0.007 The initial guess values of the variables are taken as ids = 0.5; iqs = 0.5; idr = 0.5; iqr = 0.5; vdr = 0.5; vqr = 0.5; wr = 1.0; With the values of vds, vqs, Ptot and Qtot being available, and considering the initial vds – [– rsids + ws (lssiqs + lmiqr)] = 0.8321 – 4.0175 = – 3.1854 vqs – [– rsiqs + ws (lssids + lmidr)] = – 0.1835 + 4.0225 = 3.8390 0 – (– rridr + (ws – wr) (lrriqr + lmiqs) – vdr) = 0 – (– 0.0028 – 0.5) = 0.5028 0 – (– rriqr – (ws – wr) (lrridr + lmids) – vqr) = 0 – (– 0.0028 – 0.5) = 0.5028 Ptot – (vdsids + vqsiqs + vdridr + vqriqr) = 0.5010 – 0.8243 = – 0.3233 Qtot – (vqsids – vdsiqs) = 0.0069 + 0.5078 = 0.5147 0 – [Lm(idsiqr – iqsidr) – Koptwr2] = 0 + 0.5787 = 0.5787

get the 7 ¥ 7 matrix as shown below.

ids, iqs, idr, iqr, vdr, vqr and ωr). Thus we

4.04 0 4.0 0 0 0 ˘ È -0.005 Í -4.04 -0.005 -4.0 0 0 0 0 ˙ Í ˙ 0 -0.0055 0 -1.0 0 -4.0301˙ Í 0 Í ˙ 0 0 -0.0055 0 -1.0 4.0301 ˙ Í 0 Í 0.8321 -0.1835 0.5 0.5 0.5 0.5 0 ˙ Í ˙ 0 0 0 0 0 ˙ Í-0.1835 -0.88321 ÍÎ 2.0 -2.0 -2.0 2.0 0 0 -1.1574˙˚ ids = 1.8036; iqs = – 0.4061; idr = – 1.7753; iqr = 0.6204; vdr = – 30.0477; vqr = 30.0541; wr = 8.4582; Continuing in the same way, convergence (tolerance = 0.0001) is achieved in 8 iterations. The values of the variables in iterations 2-8 are shown in Table 9.1.

Renewable Sources—Wind Energy System

Iteration

ids

iqs

idr

iqr

vdr

2

1.6337

– 0.3686

– 1.6037

0.5823

–2.9773

vqr 0.1452

wr 4.3763

3

0.8953

– 0.2058

– 0.8582

0.4170 – 1.1225

0.2777

2.3430

4

0.6164

– 0.1442

– 0.5765

0.3545 – 0.3093

0.0714

1.3710

5

0.5827

– 0.1368

– 0.5425

0.3469 – 0.0036

6

0.6007

– 0.1408

– 0.5607

0.3510

0.0408

9.9

0.0003

1.0080

– 0.0076

0.9562

7

0.6020

– 0.1411

– 0.5620

0.3513

0.0412

– 0.0075

0.9558

8

0.6020

– 0.1411

– 0.5620

0.3513

0.0412

– 0.0075

0.9558

In case of SCIG, the stator and the rotor are still described by Eqs (9.17)–(9.20). However, the rotor is short-circuited in itself and hence the voltage applied to the rotor is equal to zero, i.e., vdr = 0; vqr = 0; However, due to the absence of any converter, there is no control over the torque and the reactive power. Also, there is no question of maximum power tracking and hence power. Therefore, if the terminal (or stator) voltage and its angle, and the active zero), and Eq. (9.21) are solved to get the values of (ids, iqs, idr, iqr and ωr). Then the reactive power consumed by the SCIG can be obtained using Eq. (9.22). A 100 MW wind farm consists of 20 SCIGs connected in parallel. Each SCIG is rated at 5 MW and the detailed relevant data of the machine is Considering a system base of 100 MVA and terminal voltage of 1.0–0 the values of the stator and the rotor currents, the slip and reactive power consumed when the wind farm is supplying 90 MW. Use the Newton–Raphson method. It is to be noted that since the system base MVA is the same as the wind farm rating, the aggregated SCIG data in the system base will be the same as the individual machine data. Given vds = 1.0; vqs = 0; Ptot = 0.9 The initial guess values of the variables are taken as ids = 0.5; iqs = 0.5; idr = 0.5;

iqr = 0.5; wr = 1.0;

vds - ÈÎ- rs ids + w s (lss iqs + lm iqr )˘˚ = 1.0 - 0.5472 = 0.4528 vqs - ÈÎ- rs iqs - w s (lss ids + lm idr )˘˚ = 0 - 0.4478 = -0.4478

9.10

Computer Techniques in Power System Analysis

0 - ( - rr idr + (w s - w r ) (lrr iqr + lm iqs )) = 0 - ( -0.0033) = 0.0033 0 - ( - rr iqr - (w s - w r ) (lrr idr + lm ids )) = 0 - 0.0020 = -0.0020 Ptot - (vds ids + vqs iqs + vdr idr + vqr iqr ) = 0.9 - 0.5 = 0.4 1.0 0 0 ˘ È - 0.005 0.0993 Í- 0.0993 - 0.005 0 1. 0 0 ˙ Í ˙ The Jacobian becomes Í- 0.0053 0 0 - 0.0014 0.5 ˙ Í ˙ 0.0053 - 0.0014 0 - 0.5˙ Í 0 ÍÎ 1.0 0 0 0 0 ˙˚

ids = 0.9000; iqs = 1.1501; idr = 0.8903; iqr = 0.0951; w r = 1.0099; Convergence is achieved in 4 iterations. The values of the variables in iterations 2-4 are shown in Table 9.2.

Iteration

ids

iqs

idr

iqr

wr

2

0.9000

0.3215

0.9726

0.0910

1.0046

3

0.9000

0.3330

0.0910

0.9714

1.0051

4

0.9000

0.3330

0.9714

0.0910

1.0051

power systems with SCIG based wind farm connected at one or more bus. This is because the reactive power exchange by the SCIG is not controlled and hence cannot method of representing the wind generator in the rest of the system as negative load P and Q (of the wind farm) cannot be used. The alternate method that can be followed is described below. Suppose the wind generation system is connected to the jth bus of a n bus power system as shown in Fig. 9.4. Vdg–qdg

Idg–fdg

Vj–qj Power System Bus ‘ j’

SCIG Local Load

The power mismatch equations of bus ‘j DPj = Pjsp – Pj

Original Load

Renewable Sources—Wind Energy System

9.11

n

= Pjsp –

 (G jk cos q jk + B jk sin q jk ) |Vj||Vk| – real [(|Vj| – qj)(|Idg| – fdg)*] (9.25) k =1

DQj = Qjsp – Qj n

= Qjsp – Â (G jk sin q jk - B jk cos q jk ) |Vj||Vk| – imag [(|Vj| – qj)(|Idg| – fdg)*] (9.26) k =1

where |Idg| –fdg line and transformer, and going into bus ‘j’. This current is actually the stator current of the SCIG. I dg –fdg = i ds + jiqs

(9.27)

The terminal voltage (stator voltage) of the SCIG can be expressed as v ds + jvqs = V j –q j + j ( r + jx ) ( I dg –fdg )

(9.28)

vdr = 0; vqr = 0;), and Eq. (9.21) for the SCIG, and the equations of the ‘n’ bus system (of the form (6.48)–(6.49)) are solved simultaneously using the Newton–Raphson method to obtain the values of the stator and the rotor currents and the rotor speed (or slip) of the SCIG along with the bus voltages and angles of the remaining buses of the system. The equations (9.25)– (9.28) are used to relate the SCIG equations and the ‘n’ bus system equations while forming the mismatch equations and the Jacobian matrix.

As described in Chapter 8, the transient stability condition of a power system following a large disturbance like a fault is studied by carrying out time domain simulation δ) of the synchronous machines in the post-disturbance condition are obtained, and the maintenance/loss of synchronism involves the simultaneous solution of one set of differential equations (corresponding to the alternators) and one set of algebraic equations (corresponding to the network). Now, if some wind farm is connected to the power system, then the differential and the algebraic equations of the wind generator, the wind turbine and the rotating mass are also to be solved. Considering the two-mass model of the wind turbine and the rotating mass, they are represented by the differential equations, Eqs (9.1)–(9.3). If equations, Eqs (9.8)–(9.11). However, as mentioned in Sec. 9.2.2, the stator and rotor dynamics of the induction generator is generally neglected and they are represented by the algebraic equations, Eqs (9.17)–(9.20). The other algebraic equations of the induction generator are Eqs (9.21)–(9.22). For the single-mass model, Eq. (9.5) is used instead of Eqs (9.1)–(9.3). If SCIG is used as the wind generator, then Eqs (9.17)–(9.20) are used but with vdr = 0; vqr = 0. converters—the rotor side converter (RSC) and the grid side converter (GSC) in the

9.12

Computer Techniques in Power System Analysis

rotor circuit of the machine. The converters are connected back-to-back through a

at unity power factor, exchanging no net reactive power with the grid. Moreover, the control of the GSC is much faster than the electro-mechanical transients. Because of these reasons, generally it is not required to represent the GSC controller dynamics in a transient stability program. Instead it is represented by a current source (IGSC) which accounts for the exchange of power between the RSC and the grid. Hence, stator current and the IGSC is also neglected and it is reasonable to assume (for transient stability study) that voltage during power system disturbances. However, the RSC controls should be incorporated as these have impact on the torque/active power output and the reactive The RSC being a voltage source converter, supplies the voltage (vdr + jvqr) to voltage through Eqs (9.10)–(9.11) or Eqs (9.19)–(9.20), and thus the RSC controls ultimately the solution of the equations representing those controls will provide the required values of vdr and vqr Eqs (9.19)–(9.20). Also, the value of ωr obtained from the solution of the differential equations, Eqs (9.1)–(9.3) at every time step, is substituted in Eqs (9.19)–(9.20). After substituting the values of vdr, v and ωr, Eqs (9.17)–(9.20) turn into linear equations which can be solved to obtain the values of (ids, iqs, idr, iqr). While doing this, it should be checked that the values of idr, iqr do not exceed their respective limits. This is required from a practical point of view as the converter has a current limit. In case the computed value of idr, iqr exceeds the limit, the limiting values are taken as the value of the current for that particular time step. d-axis of the synchronously rotating voltage, thus giving rise to the relation, vds = |Vdg|; vqs = 0

(9.29)

This orientation leads to the decoupling of the d-axis control (for active power/ torque) and the q relations shown in Fig. 9.5. The reference for the rotor voltage is obtained as shown below in Eq. (9.30). The switching pulses of the RSC are obtained by pulse width modulation using this rotor voltage reference. Teref

Ê Lss ˆ ÁL y ˜ Ë m qs ¯

idrref +

P-I Control idr

vdrref

Renewable Sources—Wind Energy System

Qsref

Ê Lss ˆ ÁL V ˜ Ë m ds ¯

+

iqrref +

+

P-I Control

iqr

9.13

vqrref

yqs Lm

(

2 2 Vrref = vdr + vqr ; q r = tan -1 vqrref vdrref ref ref

)

(9.30)

However, as the dynamics of the rotor is neglected, the relation between the rotor current and the rotor voltage becomes algebraic and the P-I controller can be as the converter (rotor) current. That means

idr = idrref ; iqr = iqrref ; This assumption is reasonable since the time constant of the current controllers are usually in the range of 1 ms, which is negligible as compared to the time constant of the electromechanical transients (0.5-2.0 s). 1. The solution of the synchronous machine differential equations and the network algebraic equations provides the bus voltage magnitudes and angles. bus is also obtained. Considering the stator voltage orientation, the values of vds and vqs are obtained as in Eq. (9.29). 2. The rotor current references are obtained using the values of stator voltage, following the relationships shown in Fig. 9.5(a) and 9.5(b). 3. These rotor current references are considered to be the rotor currents and used in Eqs (9.19)–(9.20). 4. The differential equations corresponding to the wind turbine and the rotating mass, i.e., Eqs (9.1)–(9.3) are solved using the values of system parameters and the previous time-step values of the variables to obtain the current timestep values of (ωt, ωr and θtw). This value of ωr is used in Eqs (9.19)–(9.20). 5. Steps 2 and 3 convert Eqs (9.17)–(9.20) to linear algebraic equations which are then solved to get the current time step values of (ids, iqs, vdr, vqr). If the values of vdr and vqr the converter voltage rating), then they are considered to be equal to the limiting values. 6. Using the computed values of (ids, iqs, idr, iqr, vdr, vqr) and the values of vds and vqs in Eqs (9.21)–(9.22), the active and reactive power output of the Ptot and Qtot) are obtained. This Ptot and Qtot are then used for the solution of the power system network in the next time step. Similarly, the

9.14

Computer Techniques in Power System Analysis

Eqs (9.1)–(9.3) in the next time-step. This complete procedure is repeated in

bus 5 of the power system described in Example 6.2 through a transformer and a distribution line as shown in Fig. 9.3. The combined impedance of the transformer and the distribution line is (r + jx) = (0.003 + j0.02) p.u. At a particular time, the wind speed is 13.95 m/s and the wind farm injects 40 MW active power (and no reactive power) at bus 5. An additional local load of 40 MW is also connected at bus on the own base of the machine. It is considered that the synchronous machines are machines and the exciters are given in Table 9.3. A 3-phase short-circuit fault takes place in bus 4 and gets cleared after 0.20 s. Find and plot the changes in the following fault clearance: (i) Terminal bus voltage magnitude (|Vdg|) , (ii) rotor current (Ir), (iii) Active power output at the terminal (Ptot), (iv) Rotor speed (ωr). Also plot the relative rotor angle δ21 (i.e., δ1 – δ2) for the same period. Consider the rotor angle of the synchronous generator 1 as the reference. Assume that during the fault condition, the rotor is short-circuited through a resistance of 0.25 p.u. and the machine behaves like a squirrel cage induction machine.

M/C Bus xd x¢d Tdo¢ No No (p.u.) (p.u.) (sec) 1 5 1.3125 0.1813 5.89 2 6 0.8958 0.1198 6.00

xq Tqo¢ xq¢ (p.u.) (p.u.) (sec) 1.2578 0.2500 5.600 0.8645 0.1969 0.535

H (sec) 3.01 6.40

kD (p.u.) 1.0 2.5

KA

TA

20 20

0.2 0.2

(iii) Adding the additional local load at bus 5 to be the 7th bus.

Bus no. 1

Voltage Magnitude (p.u.) 1.05

Generation Angle (deg) 0.0

2

1.02

3

0.8260

–14.0886

–3.2844

4

0.9299

–9.9014

Load

MW

MVAR

MW

MVAR

48.3

23.97

0

0

25.0

11.22

0

0

0

27.5

0

0

0

0 6.5 0

Contd.

Renewable Sources—Wind Energy System

Bus no.

Voltage Magnitude (p.u.)

5

Generation Angle (deg)

9.15

Load

MW

MVAR

MW

MVAR

0

0

55.0

9.0

0.8502

–13.2274

6

0.9057

–12.5786

0

0

25.0

7

0.8533

-11.9636

0

0

– 40.13

2.5 – 0.0088

Vdg| – qdg = 0.8533 – – 11.96°.

Raphson method as shown in Example 9.1) are as given below. ids = 0.8448; iqs = – 0.0208; idr = – 0.8523; iqr = 0.2353; iqr = – 0.0114; Ptot = 0.8027; Te = 0.7244; wr = 1.1189

1.25

0.95

Vdg (p.u.)

1

Ir(p.u.)

1.3

1.2

1.15 1.1

idr = – 0.0990;

0.9

0.85

1.5

2 2.5 Time (s)

0.8

3

1.5

2 2.5 Time (s)

(a)

3

(b)

1 1.16 1.14

wr (p.u.)

Ptot(p.u.)

0.9 0.8

1.12

0.7

1.1

0.6 0.5

1.08 1.5

2 2.5 Time (s)

3

1.06

1.5

2 2.5 Time (s)

(c)

(d) r tot

3

9.16

Computer Techniques in Power System Analysis

and the machine behaves like a squirrel cage induction machine. At fault clearance, ids = 0.2531; iqs = 0.1216; idr = – 0.2558; iqr = – 0.0158; vdr = 0; vqr = 0; Ptot = 0.1079; Te = 0.1083; wr = 1.1549 The value of the rotor angles of the two synchronous machines at the point of fault clearance are d1 = 18.82°; d2 = 42.17° Time domain simulation is continued with these initial values. The plots of the following variables (i) terminal bus voltage magnitude (|Vdg|) , (ii) rotor current (Ir), (iii) active power output at the terminal (Ptot), (iv) rotor speed (ωr), and (v) relative rotor angle (δ21) of the synchronous generators are shown in Fig. 9.6 and 9.7. Variation of the relative rotor angle of synchronous generators 20

d 21(deg)

10 0 -10 -20 -30 1.5

2 2.5 Time (s)

3

δ

In this chapter we have explained the fundamental concepts underlying the simulation different components of the wind power generation system (wind turbine, generator) are discussed and suitable models to represent them are presented. Since the induction generators are mostly used as wind generators, the models of both squirrel cage and doubly-fed induction generators are discussed. The methods to incorporate the wind

power generation system in the transient stability study of large power systems is

Renewable Sources—Wind Energy System

9.17

9.1 In the system of Example 9.1, the wind farm is rated at 60 MW. Considering a wind power generation (injection at bus 5) of 30 MW and 12 MVAR, and an solution of the system. 9.2 The generator 2 of the power system described in Example 6.2 is replaced by a wind farm of 50 MW. The wind farm generates active power exactly equal to the power system. 9.3 For the power system of Example 9.1, the wind farm is considered to be SQIG

consumed by the SQIG. 9.4 to bus 4 and supplying 25 MW, there is no local load connected to bus 4 and a 3-phase short-circuit fault takes place at bus 5. Find the critical clearing time for this case.

A

Appendix

n n unknowns x1, º, xn f1(x1, x2, º, xn) = 0 f2(x1, x2, º, xn fn(x1, x2, º, xn) = 0 f1, º, fn

xi

a11x1 + a12 x2 +

+ a1n xn = u1

a21x1 + a22 x2 +

+ a2n xn = u2

an1 x1 + an2 x2 +

+ ann xn = un

More compactly we have

where A

AX = U aij X = [x1, º, xn]T and U = [u1, º, un]T

iterative methods direct methods

A.2

Computer Techniques in Power System Analysis

xi’s x1 = (u1 – a12x2 – a x – º – a1n xn)/a11 x2 = (u2 – a21x1 – a x – º – a2n xn)/a22 xn = (un – an1x1 – an2x2 – º – an, n – 1 xn – 1)/a11 Ê ˆ n Áu - a x ˜ Á i  ij j ˜ j =1 ÁË ˜¯ j πi xi = , i = 1, 2, aii

,n

aii π diagonal dominance aii

i and column i xi xi’s xi Max xi(k +1) - xi(k ) < e i

k) and (k + 1) denote the kth and (k

xi n xi(1) (i = 1, 2,

xi(k ) and

kth and (k xi(k +1) (i = 1, 2,

(

, n ) and so

, n)

xi(k +1) = uii - ai1 x1(k +1)

- ai ,i -1 xi(-k1+1) - ai ,i +1 xi(+k1)

)

- ain xn(k ) / aii , i = 1, 2,

,n

A.3

relaxation xi

Gaussian elimination method

a11x1 + a12x2 + a x = u1 a21x1 + a22x2 + a x = u2 a x1 + a x2 + a x = u a11 a21/a11 a /a11 system x1 +

a a12 u x2 + 13 x3 = 1 a11 a11 a11

(a22 - a12 a21 a11 ) x2 + (a23 - a13 a21 a11 ) x3 = u2 - u1a21 a11

(a32 - a12 a31 a11 ) x2 + (a33 - a13 a31 a11 ) x3 = u3 - u1a31 a11 x1 + b12 x2 + b13 x3 = v1 (1) (1) a22 x2 + a23 x3 = u2(1) (1) (1) a32 x2 + a33 x3 = u3(1) (1) a22 (1) (1) - a32 a22

x1 + b12x2 + b x = v1

(

)

(1) (1) (1) x2 + a23 a22 x3 = u2(1) a22

(

)

(

(1) (1) (1) (1) ˘ (1) (1) Èa33 - a23 a32 a22 ˚ x3 = u3(1) - u2(1) a32 a22 Î

x1 + b12 x2 + b13 x3 = v1

)

A.4

Computer Techniques in Power System Analysis

x2 + b23 x3 = v2 ( 2) a33 x3 = u3(2) ( 2) a33

x1 + b12 x2 + b13 x3 = v1 x2 + b23 x3 = v2 x =v ( 2) v3 = u3(2) a33

where

È1 b12 Í0 1 Í ÍÎ0 0 substitution

b13 ˘ È x1 ˘ È v1 ˘ b23 ˙ Í x2 ˙ = Ív2 ˙ ˙Í ˙ Í ˙ 1 ˙˚ ÍÎ x3 ˙˚ ÍÎ v3 ˙˚ forward backward substitution procedure

x1, x2, and x x =v x2 = v2 – b v

x1 = v1 – b12v2 – b x n akj(k -1) bkj = (k -1) akk

j ≥ k +1

(k -1) vk = uk(k -1) akk

aij(k ) = aij(k -1) - aik(k -1)bkj

i, j ≥ k + 1

ui(k ) = ui(k -1) - aik(k -1)vk

i ≥ k +1

n

xi = vi -

 bij x j ,

j =i +1

i = n, n - 1,

, 2, 1

A.5

n x1 = F1 ( x1 , x2 , …, xn ) x2 = F 2 ( x1 , x2 , …, xn ) xn = F n ( x1 , x2 , …, xn ) More compactly, let xi = F i ( x ) , i = 1, 2,

,n

xi(0) vector x(k ) = ÈÎ x1(k ) ,

, xn(k ) ˘˚

T

(

xi(k +1) = F i x1(k ) , x2(k ) ,

(

)

xi(k +1) = F i x1(k +1) , x2(k +1) , xi (i = 1, 2,

)

, xn(k ) , i = 1, 2, …, n

, xi(-k1+1) , xi(k ) , xi(+k1) , …, xn(k ) , i = 1, 2, …, n

, n)

e Max xi(k +1) - xi(k ) < e i

f (x x(0)

x

f (x) around x(0)

( ) (

) ( )

f x (0) + x - x (0) f ¢ x (0) +

(

1 x - x (0) 2!

)

2

( )

f ¢¢ x(0) +

=0

A.6

Computer Techniques in Power System Analysis

( ) (

) ( )

f x (0) + x - x (0) f ¢ x (0) = 0 x x

(1)

=x

(0)

( ) f ( x (0) )

f ¢ x (0)

-

k x

(k +1)

=x

(k )

( ) f ( x(k ) )

f ¢ x(k )

-

f